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HANCOCK'S APPLIED MECHANICS 
FOR ENGINEERS 



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THE MACMILLAN COMPANY 

NEW YORK • BOSTON • CHICAGO • DALLAS 
ATLANTA • SAN FRANCISCO 

MACMILLAN & CO., Limited 

LONDON • BOMBAY • CALCUTTA 
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THE MACMILLAN CO. OF CANADA, Ltd. 

TORONTO 



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HANCOCK'S APPLIED MECHANICS 
FOR ENGINEERS 



REVISED AND REWRITTEN 




HI G Gr S 



PROFESSOR OF THEORETICAL AND APPLIED MECHANICS IN THE 

SCHOOL OF APPLIED SCIENCE OF THE CARNEGIE 

INSTITUTE OF TECHNOLOGY 



NciD gork 

THE MACMILLAN COMPANY 

1915 

All rights reserved 



TA 






Copyright, 1909 and 1915, 
By the MACMILLAN COMPANY, 



Set up and electrotyped. Published January, 1909. 
Revised and rewritten edition, March, 1915. 




Z jf^Q 



Normooti '^xti.i 

J. S. Cashing Co. — Berwick & Smith Co. 

Norwood, Mass., U.S.A. 



MAR 13 1915 



PREFACE 

Ik the preparation of this book the author has had in 
mind the fact that the student finds much difficulty in 
seeing the applications of theory to practical problems. 
For this reason each new principle developed is followed 
by a number of applications. In many cases these are 
illustrated, and they all deal with matters that directly 
concern the engineer. It is believed that problems in 
mechanics should be practical engineering work. The 
author has endeavored to follow out this idea in writing 
the present volume. Accordingly, the title " Applied 
Mechanics for Engineers " has been given to the book. 

The book is intended as a text-book for engineering 
students of the Junior year. The subject-matter is such 
as is usually covered by the work of one semester. In 
some chapters more material is presented than can be used 
in this time. With this idea in mind, the articles in 
these chapters have been arranged so that those coming 
last may be omitted without affecting the continuity of 
the work. The book contains more problems than can 
usually be given in any one semester. 

While it is difficult to present new material in the 
matter of principles, much that is new has been intro- 
duced in the applications of these principles. The sub- 
ject of Couples is treated by representing the couples 
by means of vectors. The author claims that the chap- 
ters on Moment of Inertia, Center of Gravity, Work 
and Energy, Friction and Impact are more complete 
in theory and applications than those of any other 



vi PREFACE 

American text-book on the same subject. These are 
matters upon which the engineer frequently needs infor- 
mation ; frequent reference is, therefore, given to origi- 
nal sources of information. It is hoped that these chapters 
will be especially helpful to engineers as well as to students 
in college, and that they will receive much benefit as 
a result of looking up the references cited. In general, 
the answers to the problems have been omitted for the 
reason that students who are prepared to use this book 
should be taught to check their results and work inde- 
pendently of any printed answer. 

The author wishes to acknowledge the helpful sugges- 
tions obtained from the many standard works on me- 
chanics. An attempt has been made to give the specific 
reference to the original for material taken from engi- 
neering works or periodical literature. He wishes, more- 
over, to express his thanks to Dean C. H. Benjamin and 
Professor L. V. Ludy for their careful reading of the 
manuscript, to Professor W. K. Hatt for many of the 
problems used, and to Dean W. F. M. Goss, whose con- 
tinued interest and advice have been a constant source of 
inspiration. It is hoped that the work may be an inspira- 
tion to students of engineering. 

E. L. HANCOCK. 

Purdue University, 
November, 1908. 



PREFACE TO THE REVISED EDITION 

In the revision of this text, although rather extensive 
changes in method of treatment have been made in cer- 
tain parts, the general subject matter and order of arrange- 
ment have been, in the main, retained. The chapter on 
Dynamics of Machinery, now called Dynamics of a Rigid 
Body, has been transferred to a later position in the book 
on account of its relatively greater difficulty. 

An important change is to be found in the much larger 
use of graphical methods. The graphical and analytical 
methods have been developed simultaneously and many 
problems are given to be solved by both methods. The 
use of graphical methods has brought about the intro- 
duction of considerable new material, particularly in the 
construction of stress diagrams for trusses, and in the 
application of the equilibrium polygon to centers of grav- 
ity of plane areas, to weighted strings and linkages, and 
to bending moment diagrams. 

The problems illustrating the principles follow imme- 
diately after the development of the principles but are of 
such a nature as to require original thinking. The solu- 
tion does not consist in finding and substituting in the 
proper formula. About two hundred new problems have 
been added throughout the book. 

I wish to express my thanks to Mr. E. G. Frazer for a 
careful reading of the proof and for helpful suggestions. 

N. C. RIGGS. 
School of Applied Science, 
Carnegie Institute of Technology. 
January, 1915. 

vii 



TABLE OF CONTENTS 

CHAPTER I 
Articles 1-16 

PAGE 

Definitions 1 

Force — Inertia — Mass — Units — Vectors — Displace- 
ment — Transmissibility of Forces — Force Triangle — Force 
Polygon. 

CHAPTER II 

Articles 17-24 

Concurrent Forces 10 

Conditions for Equilibrium — Equilibrium of Three 
Forces — Concurrent Forces in Space — Moments — Vari- 
gnon's Theorem. 

CHAPTER III 

Articles 25-32 

Parallel Forces 27 

Moment of Resultant of Parallel Forces — Center of Par- 
allel Forces. 

CHAPTER IV 

Articles 33-43 

Center of Gravity 38 

Center of Gravity by Integration — Graphical Methods — 
Simpson's Rule — Durand's Rule — Theorem of Pappus. 

ix 



X TABLE OF CONTENTS 

CHAPTER V 
Articles 44-53 

PAGE 

Couples 71 

Definition — Moment of a Couple — Combination of 
Couples — Equivalent Couples — Vector Representation of 
Couples. 

CHAPTER VI 

Articles 54-59 

Non-concurrent Forces 82 

Forces in a Plane — Conditions of Equilibrium — Forces in 
Space — Graphical JMethocl for Forces in One Plane — Equi- 
librium Polygon — Stresses in Frames — Bow's Notation. 

CHAPTER VII 

Articles 60-84 

Moment of Inertia 106 

Definition of Moment of Inertia — Radius of Gyration — 
Parallel Axes — Inclined Axes — Product of Inertia — Prin- 
cipal Axes — Graphical Method — Use of Simpson's Rule — 
Ellipse of Inertia — Ellipsoid of Inertia. 

CHAPTER VIII 

Articles 85-101 

Flexible Cords . 148 

Introduction — Suspended Weights — Graphical Methods 
— String Polygon — The Linked Arch — Masonry Arch — 
Bending Moments — Tensile Stresses in Beams — Shear — 
Cords and Pulleys — Weighted Cord as Parabola — Cate- 
nary — Hyperbolic Functions. 



TABLE OF CONTENTS xi 

CHAPTER IX 

Articles 102-114 

PAGE 

Motion in a Straight Line 183 

Velocity — Acceleration — Xewton's Laws of Motion — 
Harmonic Motion — Motion under Different Laws of Force 

— Relative Velocity. 

CHAPTER X 
Articles 115-124 

Curvilinear Motion 204 

Velocity — Acceleration — Centrifugal Force — Motion in 
a Vertical Plane — Simple Pendulum — Cycloidal Pendulum 

— Projectiles — Motion in a Twisted Curve. 

CHAPTER XI 

Articles 125-133 

Rotary Motion 232 

Angular Velocity and Angular Acceleration — Plane Mo- 
tion of a Body — Instantaneous Axis of Rotation — Combi- 
nation of Simultaneous Rotations — Foucault's Pendulum. 



CHAPTER XII 

Articles 134-153 

Work and Energy 245 

Definitions — Friction Forces — Graphical Representation 
of Work — Conservation of Energy — Pile Driver — Steam 
Hammer — Work of Impressed Forces — Brake-shoe Test- 
ing Machine — Kinetic Energy of Rolling Bodies. 



XU TABLE OF CONTENTS 

CHAPTER XIII 
Articles 154-176 

PAGE 

Friction 283 

Coefficient of Friction — Laws of Friction for Dry Sur- 
faces — Friction of Lubricated Surfaces — The Wedge — 
Method of Testing Lubricants — Rolling Friction — Anti- 
friction Wheels — Resistance of Roads — Roller Bearings — 
Ball Bearings — Friction Gears — Friction of Belts — Power 
Transmitted by a Belt — Transmission Dynamometer — 
Creeping of Belts — Friction of a Worn Bearing — Friction 
of Pivots — Absorption Dynamometer — Friction Brake — 
Prony Friction Brake — Friction of Brake Shoes — Train 
Resistance. 

CHAPTER XIV 

Articles 177-206 

Dynamics of Rigid Bodies 334 

D'Alembert's Principle — Translation of a Rigid Body — 
Rotation about a Fixed Axis — Equations of Motion of a 
Rotating Body — Reactions of Supports of a Rotating Body 

— Compound Pendulum — Centers of Oscillation and Sus- 
pension — Experimental Determination of Moment of In- 
ertia — Determination of ^ — The Torsion Balance — 
Rotating Body under the Action of No Forces — Rotation 
of a Body about a Fixed Axis with Constant Angular 
Velocity — Rotation of Locomotive Drive Wheel — Stand- 
ing and Running Balance of a Shaft — Rotation of Fly- 
wheel of Steam Engine — Rotation and Translation — The 
Connecting Rod — Angular Momentum — Torque and An- 
gular Momentum — Moment of Momentum of a Body with 
One Point Fixed — Vector Representation of Angular Mo- 
mentum — Kinetic Energy of a Body with One Fixed Point 

— Vector Rate of Change Due to Rotation — Rate of Change 
of Angular Momentum of a Body Due to Rotating Axes — 



TABLE OF CONTENTS xiii 



Motion of Center of Gravity of a Body — Gyroscope, Hori- 
zontal Axis — The Gyroscopic Couple — Car on Single Rail 
— Gyroscope, Inclined Axis. 



CHAPTER XV 

Articles 207-215 

Impact 390 

Definitions — Direct Central Impact — Momentum and 
Kinetic Energy in Impact — Elasticity of Material — Im- 
pact Tension and Impact Compression — Direct Eccentric 
Impact — Centers of Percussion and Instantaneous Rotation 
— Oblique Impact of Body against Smooth Plane — Impact 
of Rotating Body. 



APPLIED MECHANICS FOR ENGINEERS 

CHAPTER I 
DEFINITIONS 

1. Introduction. — The study of the subject of mechanics 
involves a study of matter^ space, and time, and of the be- 
havior of bodies under the action of forces. The subject 
as presented in tliis book consists of two parts ; viz. 
statics, including the study of bodies under the action 
of systems of forces that are in equilibrium (balanced), 
and dynamics, including a study of the motion of bodies. 

2. Force. — A body acted upon by the attraction or 
repulsion of another body is said to be subjected to an 
attractive or repulsive force, as the case may be. In sim- 
ple terms a force is a push or a pull. Forces are usually 
defined by the effects produced by them, as, for example, 
we say, a force is something that produces motion or 
tends to produce motion, or changes or tends to change 
motion, or that changes the size or shape of a body. 
Forces always occur in pairs ; for example, a book held 
in the outstretched hand exerts a downward pressure on 
the hand, and the hand exerts an equal upward pressure 
on the book. 

3. Inertia. Mass. — It is a matter of common experience 
that bodies vary in the amount of resistance that they 
offer to a change of their state of rest or motion. Thus 



2 APPLIED MECHANICS FOR ENGINEERS 

it is more difficult to stop a swiftly moving ball of iron 
than one of wood having the same dimensions and speed. 
It is easier to set in motion the wooden ball than to give 
the same speed to the iron ball of the same size. 

The property of resistance to change of its state of rest 
or motion that every body has is called its inertia. 

The word mass is used in the sense of a measure of inertia. 

4. The Unit of Mass. — A certain arbitrary piece of 
platinum carefully preserved by the British government 
is known as the standard mass of one pound avoirdupois. 
Any other piece of matter which under the action of a 
certain force has its motion changed in the same way as 
the standard mass of one pound under the same condi- 
tions is also called a mass of one pound. 

The corresponding unit in the French system is the gram. 

5. The Unit of Force. — The pull, or attraction, that the 
earth exerts upon a mass of one pound at sea level and 
latitude 45° is called the force of one pound. Similarly, 
the force of one gram is the force that the earth exerts 
upon a mass of one gram at sea level and latitude 45°. 
These units of force are called the gravitational units of force. 

The poundal is defined as that force which acting alone 
on a mass of one pound for one second produces in that 
mass a change of velocity of one foot per second. 

The dyne is defined as that force which acting alone on 
a mass of one gram for one second produces in that mass 
a change of velocity of one centimeter per second. 

The poundal and dyne are called absolute units of force. 

The force with which the earth attracts a body is called 



DEFINITIONS 3 

the weight of the body. Mass differs from weight, in that 
the weight varies with the position on the surface of the 
earth and with the height above the surface, while the 
mass remains the same. The weight of a body may be 
determined by means of the spring balance. The equal- 
armed balance gives the same weight regardless of dis- 
tance from the center of the earth. The equal-armed 
balance really measures mass. 

6. Unit Weight. — The weight of a cubic foot of a sub- 
stance will be called the unit weight of the substance and 
will be represented by 7. Below is given a table of such 



TABLE I 

Unit Weights and Specific Gravity of Some Materials 

(Kent's " Engineer's Pocket Book ") 



Material 


Specific Gkavitt 


Unit Weight 


Brick 






Soft 


1.6 


100 


Hard 


2.0 


125 


Fire 


2.24-2.4 


140-150 


Brickwork — mortar 


1.6 


100 


Brickwork — cement 


1.79 


112 


Concrete 


1.92-2.24 


120-140 


Copper 


8.85 


552 


Earth — loose 


1.15-1.28 


72-80 


Earth — rammed 


1.44-1.76 


90-110 


Iron — cast 


7.21 


450 


Iron — wrought 


7.7 


480 


Masonry — dressed 


2.24-2.88 


140-180 


Pine — white 


.45 


28 


Pine — yellow 


.61 


38 


Steel 


7.85 


490 


White Oak 


■ .77 


48 



4 APPLIED MECHANICS FOR ENGINEERS 

weights at sea level. The unit weight of a substance 
divided by the unit weight of pure water gives its specific 
gravity. 

7. Rigid Body. — In studying the state of motion or 
rest of a body due to the application of forces acting 
upon it, the deformation of the body itself, due to the 
forces, may be disregarded. When so considered, it is 
customary to say that the body is a rigid body. Unless 
otherwise stated bodies will be considered as rigid bodies 
in this book. 

8. Vectors. — Any quantity that has magnitude and 
direction may be represented graphically by a directed 
segment of a straight line. The length of the segment 
is taken to measure the magnitude of the quantity, and 
the direction of the segment to indicate the direction of 
the quantity. 

The directed segment of the line is called a vector^ and 
the quantity it represents, a vector quantity. 

9. Displacement. — By the displacement of a body is 
meant its change from one position to another. 

Since a change of position involves magnitude and direc- 
tion, a displacement may be represented by a vector, the 
length of the vector representing the distance from the first 
position to the second and the direction of the vector repre- 
senting the direction of the ,second position from the first. 

From the definition of a displacement it follows that 
two successive displacements are equivalent to a single 
displacement. Thus, if a man walks due east one mile 
and then due north one mile, we might represent his dis- 



DEFINITIONS 5 

placement from the original position by a vector drawn 
northeast to represent a length equal to V2 miles. Or, 
in Fig. 1, if P^ represents a displacement of a body in 
the direction indicated and Pg ^ subsequent displacement 
in the direction of Pg' then H represents a displacement 
equivalent, to Pj and P^. It is seen that H may be deter- 
mined by constructing a parallelogram on P^ and P^ as 
sides and drawing the diagonal. 




P. 

Fig. 1 

10. Representation of Forces. — A force has a certain 
magnitude, acts in a certain direction, and has a definite 
point of application. If a man, for example, attaches a 
rope to a log and pulls on the rope, his pull may be meas- 
ured in pounds, it acts along the rope, and its point of 
application is the point of attachment of the rope to the 
log. It is convenient, for the purpose of analysis, to 
represent forces by vectors, the length of the vector 
representing the magnitude of the force and its direction 
giving the direction in which the force acts. Thus, a 
10-lb. force, acting in a direction 30° with the horizontal, is 
represented by a vector drawn in the same direction and 
having its point of application in the body, and having a 
length representing 10 lb. (In this case, if 1 in. represents 
2 lb., the length of the vector is 5 in.) The line along 
which a force acts will be referred to as its line of action. 



6 APPLIED MECHANICS FOR ENGINEERS 

11. Transmissibility of Forces. — It is a matter of experi- 
ence that the point of application of a force may be changed 
to any point along its line of action without changing the 
effect of the force upon the rigid body. This, of course, is on 
the assumption that all the force is transmitted to the body. 
The law may be stated as follows : The point of application 
of a force may he transferred anywhere along its line of action 
without changing its effect upon the body upon which it acts. 

12. Concurrent Forces. — When the lines of action of 
two or more forces pass through the same point, the 
forces are known as concurrent forces. 

13. Resultant of Two Concurrent Forces. — If two con- 
current forces act on a body, there is some single force 
that might be applied at the point of intersection of the 
forces to produce the same effect. This single force is 
called the resultant of the two forces. Experiment shows 
that it may be found as follows : construct upon the vec- 
tors representing the forces, laid off from the intersection 
of their lines of action, a parallelogram and draw the 
diagonal from the point of application. This diagonal 
represents the resultant of the two forces in magnitude 
and direction and it is the line of action of the resultant. 




Thus, if Pj and P^ (Fig. 2) are the forces, then It is 
the resultant. 



DEFINITIONS 



Algebraically B = VjPiS + r^2^2 P1JP2 cos AOB. 

Instead of speaking of the vector which represents a 
force, we shall for the sake of brevity speak of the vector 
as the force. 

14. Resolution of Force. — We have just seen how two 
concurrent forces may be replaced by a single force called 
their resultant. In a similar way a single force may be 
resolved into two forces. These forces are the sides of a 
parallelogram of which the single force is a diagonal. It 
is clear, then, that there are an infinite number of compo- 
nents into which a single resultant may be resolved. It 
is necessary, therefore, in speaking of the components of 
a force, to state specifically which are intended. It will 
be seen in problems that follow that the components most 
often used are at right angles to each other, and are 
usually the vertical Siud horizontal components. In such a 
case the components are the projections of the force on the 
vertical and horizontal lines. 

15. Force Triangle. — It follows directly from the paral- 
lelogram law of forces that if we draw from any point a line 
parallel and equal to one of two concurrent forces, P^ ^^7' 
and from the extremity of this line another line parallel 
and equal to P^. then the remaining side of the triangle 
will represent the resultant R. This triangle is called 
the force triangle. In general, the resultant of two con- 
current forces may be found by drawing lines parallel to 
the forces as above. The line necessary to complete the 
triangle is the resultant, and its direction is always away 
from the point of application. The equal and opposite 



8 



APPLIED MECHANICS FOli EN GIN EE US 



of this resultant would be a single force that would hold 
the two concurrent forces in equilibrium. 

The single force which will balance a given set of forces 
is called their equilihrant. 

16. Force Polygon. — If more than two forces are con- 
current, we may find their resultant by proceeding in a 
way similar to that outlined above. Thus, let the forces 
be P^^ Pg' -^3' -^4' ®t^* (-^^^* ^)' ^^^ passing through a 

P. 





Fig. 3 

point ; from any point draw a line equal and parallel to Pj, 
from the extremity of the line draw another equal and 
parallel to P^, from the extremity of this last line draw 
another equal and parallel to Pg, and proceed in the same 
way for the other forces. The figure produced will be a 
polygon whose sides are equal and parallel to the forces. 
The resultant of the given forces is then represented in 
magnitude and direction by the line necessary to close the 
polygon, and its line of action passes through the intersec- 
tion of the given concurrent forces. 

The arrow, representing the direction of the resultant, 
will always be away from the point of application. (See 
Fig. 8.) 

By drawing the lines OA^ OB, 00, etc., it is easy to 



DEFINITIONS 9 

see that OA represents the resultant of P-^ and P^^ that 
OB represents the resultant of OA and Pg, and so of P^, 
Pgi ^^^ P^^ 6tc. That is, the force polygon follows 
directly from the force triangle. If the polygon be closed, 
the system of forces will be in equilibrium, and con- 
versely. The single force necessary to produce equi- 
librium will, in any case, be equal and opposite to the 
resultant R. The student should construct force polygons 
by taking the forces in different orders and checking the 
resultant in each case. 

By means of the force polygon it is easy to find graph- 
ically the resultant of any number of concurrent forces 
in a plane. The work, however, must be done accurately. 

The student should show that the force polygon may 
be used for finding the resultant of concurring forces in 
space, by considering two forces at a time. The force 
polygon in this case is called a twisted polygon. 

Problem 1. Find graphically the resultant in magnitude, direc- 
tion, and point of appjication of the following four concurrent forces 
in one plane: 80 lb. in direction E., 60 lb. E. 50° N., 100 lb. W. 
40° N., and 120 lb. E. 30° S. Check by taking the forces in different 
orders. 

Problem 2. Find graphically the resultant of the following four 
concurrent forces : (a) 50 lb. directed E. in a horizontal plane, 
{h) 80 lb. directed N. 40° W. in the horizontal plane, (c) 100 lb. 
directed 30° above the horizontal, the projection of the force on the 
horizontal plane being directed S. 20° E., and {d) 70 lb. directed 60° 
above the horizontal, the projection of the force on the horizontal 
plane being directed W. 40° S. 

Suggestion. Resolve the forces (c) and (ri) into their horizontal 
and vertical components before combining with the other forces. 



CHAPTER II 



CONCURRENT FORCES 



17. Concurrent Forces in a Plane. — It will often be 
convenient to consider forces as acting on a material 
point ; this is equivalent to considering the mass of the 

body as concen- 
trated at a point. 
Given a material 
point (Fig. 4) 
acted upon by a 
number of forces 
in a plane, P^^ 

2' 3' 4' etc., 
making angles 

^1^ ^^2' ^3' ^4' ®tC., 

respectively, 
with the posi- 
tive a:-axis, it is 
desired to find 
the resultant of 
all of them in magnitude and direction ; that is, the single 
ideal force that would produce the same effect as the sys- 
tem of forces. 

Each force P may be resolved into components along 
the X- and ^-axes, giving P cos a along the 2;-axis, and 

10 




CONCURRENT FORCES 



11 



P sin a along the ?/-axis. The sum of these components 
along the .-r-axis may be expressed, 

Sjr= Pj cos OCj + /*2 ^OS "2 "^ -^3 ^^^ ^3 "^" 6tC., 

the proper algebraic sign being given cos a in each case. 
In a similar way the sum of the components along the 
y-axis may be written, 

^Y= P^ sin a^ + P^ sin a^ + P^ sin a^ + etc. 

These forces, SX and 2 Z", may now replace the original 
system as shown in Fig. 5 ; and these may be combined 

Y 




into a single resultant which is the diagonal of the rec- 
tangle of which the two forces are sides (Art. 15). This 
gives the resultant in magnitude and direction, and this 
resultant force is the single force which, if allowed to act 
upon the material point, would produce the same effect as 



12 



APPLIED MECHANICS FOB ENGINEERS 



the system of forces. Analytically the resultant may be 

expressed, ^ ^ V(2X)H(ST?, 

and its direction makes an angle a with the a;-axis such 



that tan a — 



^X 



(See Fig. 5.) If the forces are in 



equilibrium, this resultant force must be equal to zero; 
that is, V(2X)2 + (2F)^ = 0. 

This means that (^Xy + (2 Y^ = 0, that is, that the sum 
of two squares must be zero; but this can happen only 
when each one, separately, is zero (since the square of a 
real quantity cannot be negative). We therefore have as 
the necessary and sufficient conditions for the equilibrium 
of a material point, acted upon by a system of concurring 
forces in a plane, ^X = 0, aud 2 r := 0. 

When II is not zero, the system of forces causes accel- 
erated motion in the direction of i?; when i^ = 0, the 

material point remains at 
rest, if at rest, or continues 
in motion with uniform veloc- 
ity, if in motion. In this 
case the system of forces is 
said to be balanced. 

As an illustration of the 
foregoing, consider the case 
of a body of weight Gr situ- 
ated on an inclined plane, 
making an angle with the 
horizontal. (See Fig. 6.) Tliere is a certain force P, 
making an angle <p with the plane, whose component along 




Fig. G 



CONCURRENT FORCES 



la 



the plane acts upwards, and also a force of friction F 
upwards. The other forces acting on the body are (7, 
the force of gravity acting vertically, and iV, the normal 
pressure of the plane. Taking the a;-axis along the plane 
positive upward and the ^-axis perpendicular to it positive 
upward, we get, 

SX=Pcos(/) + ^- a sine, 
and 2F=i\^+ P sin (/)-G^ cos ^. 

For equilibrium 

Pcos^ + P- asin^ = 0, 
iV"+ P sin </) - (7 cos l9 = 0. 

Therefore, N = G- cos — P sin <^, 
a sin O-F 



P = 



cos (f) 



This last equation gives the magnitude of P required to 
preserve equilibrium, supposing that the force of friction, 
0, and <f) are known. 

Problem 3. Given three concurring forces, 100 lb., 50 lb., and 
200 lb., whose directions referred to the x-axis are 0"^, 60°, 180°, 
respectively ; find the resultant in 
magnitude and direction. 

Problem 4. A body (Fig. 7) whose 
weight is G is drawn up the inclined 
plane with uniform velocity due to 
the action of the forces P and P'. 
Find the force of friction F, and the 
normal pressure iV, if P = 100 lb., 
P'=100 lb., 6^=160 lb. P acts 
parallel to the plane and P' acts hori- 
zontally. Fig. 7 




14 



APPLIED MECHANICS FOR ENGINEERS 



18. A Body in Equilibrium under the Action of Two Forces. 
Tension, Compression. — Two forces are in equilibrium when 
and only when their lines of action coincide and one of the 
forces is equal in value but opposite in direction to the 

other. In roof and 

^ ' ^ ^> bridge trusses, in cranes, 

and in other jointed 
structures all forces act- 




<-^-€: 



(a) 



ih) 



Fig. 8 



ing on each member are often considered as applied at two 
points near the ends of that member, and the member may 
thus be regarded as acted upon by only two forces. Where 
such is the case the two forces acting upon the member 
must be equal and opposite and their lines of action must 
coincide with the line joining the points of application of 
the forces. 

If the two forces act towards each other, the member is 
said to be in compression. If the forces act away from 
each other, the member is said to be in tension. Thus in 
Fig. 8 the member is in compression in (a) and in tension 
in (6), the amount of compression or tension being P. 

B 




Fig. 9 



Illustration. In Fig. 9 a horizontal pin carrying the 
weight IT passes through the members AB and BC which 



CONCURRENT FORCES 



15 



are inclined respectively at angles 60° and 30° to the 
horizontal and held at A and by horizontal pins perpen- 
dicular to the plane ABC. If the weights of AB and BO 
are small compared to TT, AB and BO may each be re- 
garded as acted upon by only two forces. The values of 
these forces may be found as follows: Since AB and BO 
are in equilibrium under the action of only two forces, 
those acting at the pins, the forces acting on each member 




must be along the lines joining the pins. The members 
themselves must then exert on the pins forces equal and 
opposite to those that the pins exert on the members. 
Hence acting on the pin at B^ holding it in equilibrium, 
are the forces TF", P, and §, acting respectively downward, 
along AB^ and along OB (Fig. 10). Applying the 
conditions for equilibrium, 2X =0, 2 Z = 0, there results 
P cos 60° - Q cos 30° = 0, 
F sin 60° + Q sin 30° - W= 0. 
Solving these equations, we find 

P=.866 W, 



16 



APPLIED MECHANICS FOE ENGINEERS 



A graphical solution is obtained by laying out the force 
triangle for the forces in equilibrium at the point B and 
measuring the sides representing P and Q (Fig. 11). 

Problem 5. A weight of 10 ions is supported as shown in Fig. 
12. Find the force acting in the tie A and the member B. 





Problem 6. Two ropes of length 5 ft. and 8 ft. are attached at 
one end to a weight of 500 lb. The other ends of the rope are at- 
tached to two points 9 ft. apart on a horizontal line. When the 
weight is suspended by the ropes, find the tension in each rope. 

19. Body in Equilibrium under the Action of Three Forces. 

— ^ three forces are in equilibrium^ any one of them must be 
the equilibrant of the other two. Its line of action inust 
therefore pass through the intersection of the lines of action 
of the other two. Also the vector of the third force must 
be equal and opposite to the diagonal of the parallelogram 
formed on the vectors of the other two forces as sides. 
Imposing these conditions on any body in equilibrium 
under the action of three forces will usuall}^ suggest an 
easy method of finding two of the forces acting on the 
body when the third one is known. 



CONCURRENT FORCES 



17 



Illustration. A gate 9 ft. wide bj 5 ft. high, weigh- 
ing 80 lb., is hung by two hinges distant respec- 
tively 6 in. from the top and the bottom of the gate. 
If the lower hinge carries all of the weight (i.e. the upper 
hinge exerts no upward force 
on the gate), find the reactions 
of the hinges. The weight of 
the gate may be assumed to 
act at the center. 



n 




Solution. The gate is acted ^^^- ^^ 

upon by three forces : the weight, 80 lb., acting down- 
ward at the center, a horizontal force exerted by the upper 
hinge, and a force unknown in magnitude and direction 
exerted by the lower hinge. These three forces must pass 

through a common point, the 
point A., Fig. 14, and form a 
closed triangle of forces. 
Hence, if AB = 80 lb., the forces 
P and Q in Fig. 14 represent on 
the same scale the reactions of 
the lower and upper hinges re- 
spectively. Tlie values may be found by measurement, 
or by solving the triangle ABC. Comparing the similar 
triangles ABO and AED, one a triangle of forces, the 
other a triangle of distances, we have 




Fio. 14 



Q ^ 4.5 
80 4 



.-. § = 90 lb. 



80 4 



.-. P = 120.4 lb. 



18 



APPLIED MECHANICS FOR ENGINEERS 



Problems of this nature are more easily solved 
of the theory of moments developed later. 

Problem 7, Neglecting the weight of the members AD 
in Fig. 15, find analytically and graphically the stress in BC 
reaction at ^. 

Suggestion. J5C is acted upon by two forces, one at B 
at C. AD is acted upon by three forces. For the auab^t 
first find the values of ^0 and BO and compare the - . 
with the triangle A OB. 



by use 



and BC 
and the 

and one 




2000 LBS 




Fig. 15 



Fig. 16 



Problem 8. A post 20 ft. high is hinged at the foot and stands 
vertically. Two ropes ^C and ED on opposite sides of the post and 
in a plane perpendicular to the axis of the hinge make angles of 50° 
and 40° respectively with the horizontal and are attached to the post 
at distances of 10 ft. and 20 ft. respectively from the bottom. What 
tension in ED will cause a tension of 500 lb. in BC, and what will 
then be the reaction at the foot of the post? Solve graphically. 

Problem 9. Neglecting the weight of the truss in Fig. 16, find 
analytically and graphically the reaction at A and the tension in BC. 

Problem 10. Find analytically and graphically the value of the 
horizontal force P that will just raise the corner A of the block in 
Fig. 17 from the floor. Find also, graphically, the value of P when 
A is raised 6 in. from the floor. When A is raised 1 ft. 



CONCURRENT FORCES 



19 



Problem 11. The column AB, Fig. 18, is one foot square, 15 ft. 
long, and weighs 1200 lb. Find graphically the tension in the rope 
and the reaction at A when the ang-Ie which the column makes with 



f 

41 



m' 




A\ 
_ Fig. 17 Fig. 18 

the horizontal is 0°, 30°, 60°. The pulley at C is small. Consider 
the weight of the column as acting at its center. 

Problem 12. A wheel is about to roll over an obstruction. The 
diameter of the wheel (Fig. 19) is 3' and its weight 800 lb. Find the 
horizontal force P through the center necessary to start the wheel 
over the obstruction. 




Fig. 19 

Problem 13, An angle iron, whose 
weight is 20 lb. and angle a right angle, 
rests upon a circular shaft, radius 2 in. 
Find the normal pressure at A and B (Fig. 
20). 




Fig. 20 



20. Concurrent Forces in Space. — Let P^, Pg^ ^3^ ®^^' ^® 
a set of concurrent forces not in one plane, and let their 



20 



APPLIED MECHANICS FOR ENGINEERS 



direction angles be respectively Wj, /3j, 7^ ; a^, ^^, y^ ; ccg, 
/Sg, 73 ; etc. (The direction angles of a force are the 




Fig. 21 

angles at the point of application of the force measured 
from the positive direction of the coordinate axes to the 
vector representing the force.) The force P^ may be re- 
solved into components parallel to the coordinate axes, 
X^ = Pj cos ocj, Y^ = P^ cos y8j, Z^ = Pj cos 7^. 

The resultant force may be found in magnitude and 
direction by an analysis similar to that used in Art. 17. 
The sum of the components of all the forces parallel to 
the 2:-axis is 

2X= Pj cos cjj -\- Pg cos 0^2 + P3 cos ctg + etc., 
the sura of the components parallel to the ?/-axis, 

Sy = Pj cos /Sj + Pg cos 13^ + Pg cos ySg + etc., 
and the sum of the components parallel to the si-axis, 
2Z= Pj cos 7j 4- P2 cos 72 + Pg cos 73 + etc. 
The original system of forces may now be replaced by 
a system of three rectangular forces 2X, 2 F, and 2Z 




CONCURRENT FORCES 21 

(Fig. 22). Finally, this system may be replaced by a 
resultant which is the diagonal of a parallelopiped con- 
structed with 2X, 2P, ^Z as edges. 
In magnitude this resultant may be 
expressed 

i^= V(2X)2+ (2F)2+ (SZ)2 

(see Fig. 22) and its direction given by ^ig. 22 

the angles ot, /3, and 7. These angles are given by the 

equations 

cos a = -— — , cos p = -— — , cos 7 = 

For equilibrium R must be ; that is, 

(sx)2 + (sr)2+(sz)2 = o, 

and therefore. 

This gives three equations of condition from which three 
unknown quantities may be determined. In the preced- 
ing case of Art. 17 there were only two equations of 
condition 2X= and 2 1^= 0; consequently, only two un- 
known quantities could be determined. 

Problem 14. Prove that if ct, p, y are the direction angles of any 
straight line, then cos^ a + cos^ /? + cos^ y = 1. 

Problem 15. Three men (Fig. 23) are each pulling with a force 
P at the points a, &, and c, respectively. What weight Q can they raise 
with uniform motion if each man pulls 100 lb. ? Each force makes 
an angle of 60° with the horizontal and the projections of the forces 
on a horizontal plane make angles of 120^ with each other. Solve 
analytically and also graphically by projecting each force on the 
vertical and horizontal planes. 



22 



APPLIED MECHANICS FOR ENGINEEliS 




Problem 16. Three concurring forces act 
upon a rigid body. Find the resultant in mag- 
nitude and direction. The forces are defined 
as follows : 



72=-^ 



Pi = 75 lb. ; a, = 63° 27' ; /3, = 48° 36' ; y^ 

P, = 80 lb. ; a^ = 153° 44' ; /S^ = 67° 13' 

Ps = 95 lb. ; «3 = 76° 14' ; ^3 = 147° 2' ; y., = ? 

Hint, y^, y^, and yg may be found from 
the relation : 

cos^ a + cos^ /3 + cos- y = 1. 

Problem 17. Each leg of a pair of shears 

(Fig. 24) is 50 ft. long. They are spread 20 

Fig. 23 ft. at the foot. The back stay is 75 ft. long. 

Find the forces acting on each member when lifting a load of 20 tons 

at a distance of 20 ft. from the foot of the shear legs, neglecting the 

weight of the structure. 




20 TONS 



Fig. 24 



21. Moment of a Force. — The moment of a force with 
respect to any point is defined as the product of the force 
and a perpendicular from the point to the line of action 



CONCURRENT FORCES 23 

0/ the force. Let P (Fig. 25) be the force and the 
point and a the perpendicular distance of the force from 
the point ; then Pa is the moment of the force with 
respect to the point 0. This moment is measured in 
terms of the units of both force and lengthy viz. pound- 
feet or pound-inches, and is read pound-feet or pound- 
inches to distinguish it from foot-pounds of work or 
inch-pounds of work. 

For convenience the algebraic sign of the moment is said 
to be positive when the moment tends to turn the body in a 
(direction counter-clockwise^ and 
negative when it tends to turn the 
body in the clockwise direction. 

The moment may be repre- 
sented geometrically as follows : 
let EF represent the magnitude of 
P, drawn to the desired scale, and 
draw EO and FO. The area of 
the triangle OFF = ^ EFa, or 

EFa = 2 A OFF ; that is, the moment of the force with 
respect to a point is geometrically represented hy twice the 
area of the triangle., whose base is the line representing the 
magnitude of the force and whose vertex is the given point. 

If the moment of the force is negative, then the moment 
is represented by minus twice the area of the triangle. 

22. Varignon's Theorem of Moments. — The moment of the 
resultant of two concurring forces with respect to any point 
in their plane is equal to the algebraic sum of the moments 
of the two forces with respect to the same point. 




24 



APPLIED MECHANICS FOR ENGINEERS 



Let P and Q be any two concurrent forces and a 
point in tlieir plane. Through draw a line parallel to 
the line of action of one of the forces, as P. Let the seg- 
ment AC cut off on the line of Q by this line be taken to 

D CO D 




-1 (a) B A ^^j ^B 

Fig. 26 

represent the force Q, and let AB represent the force P 
to the same scale. Then, using the upper sign for case (a) 
and the lower for case (i). Fig. 26, 

mom P + mom Q = 2 (area of OAB ± area of OAC). 
But area of OAB = area of ABB = area oi AOB. 

.'. 2 (area of OAB ± area of OAC) = 2 area of OAD 

= mom B. 
.*. mom P + mom Q = mom B. 

23. Moment of a Force with Respect to a Line. — Let P be 

any force and AB any line (Fig. 27). The moment of P 

with respect to AB is defined 
as follows : Resolve P into 
two components one of which 
is parallel to, and the other 
perpendicular to, AB. The 
product of the component per- 
pendicular to AB and the per- 
pendicular distance of thiscom- 
YiG, 27 ponent from AB is called the 




CONCURRENT FORCES 



25 



moment of P with respect to AB. (In Fig. 27 the mo- 
ment of P with respect to AB is aPy} More briefly the 
definition is : The moment of a force with respect to a 
line is the moment of the projection of the force upon a 
plane perpendicular to the line, with respect to the inter- 
section of the line and the plane. The moment is con- 
sidered positive or negative according as the tendency to 
rotatiori. by the force is counter-clockwise or clockwise. 
The sign of the moment will then change if the observer 
changes from one side of the plane on which the projection 
is made to the other, and in comparing the moments of 
several forces with respect to a line the forces should all 
be projected upon the same plane and their projections 
viewed from the same side of that plane. 

24. Moment of the Resultant of Two Concurrent Forces with 
Respect to a Line. — The sum of the moments of two con- 
current forces with respect to any line is equal to the moment 
of their resultant with respect to that line. 

Proof : Let P and Q be any two concurrent forces, B 
their resultant, and AB j/r 
any line (Fig. 28). 
Through the intersection 
of P and Q pass a plane 
MN perpendicular to AB^ 
cutting ^^in 0. Project 
P, ft and B on MJSf, the 
projections being respec- 
tively Pj, Qy, and By Fig. 28 
From the definition of projection it follows at once thnt 
Pj is the resultant of P^ and Qy The moments of P, Q, 




26 APPLIED MECHANICS FOB ENGINEERS 

and M with respect to AB are resf)ectively equal to the 
moments of Pj, Q^^ and R^ with respect to 0. (Definition.) 
By Art. 22, with respect to 0, 

mom Pj + mom §j = mom My 

Therefore, with respect to AB^ 

mom P + mom Q = mom i?. 

If there are three or more forces, the application of the 
above theorem to the resultant of two of the forces and a 
third force, etc., proves that the sum of the m,oments of 
any number of concurre7it forces with respect to any line is 
equal to the moment of their resultayit with respect to that 
line. 

CoKOLLARY. From the definition of the moment of a 
force with respect to a line it follows that the sum of the 
moments of two forces of equal numerical value but oppo- 
site in direction and acting in the same line is zero. 

Use will be made of these principles in a later chapter. 



CHAPTER III 



PARALLEL FORCES 



25. The Resultant of Two Parallel Forces. — In consid- 
ering two parallel forces three cases arise : (a) when the 
forces are in the same direction ; (5) when they are 
unequal and in opposite directions ; (c*) when they are 
equal and in opposite directions, but have different lines 
of action. 

In case (c) the two forces form a couple. It will be 
shown later that there is no single force that will replace 
them. 

In cases (a) and (5) the two forces have a resultant. 
Its value and line 
of action are found 
as follows : 

Let the two 
forces be Pj and 
Pgi acting at the 
points A^ and A^. 
At A^ and A^ in 
the line A^A^ put 
in two equal and Fig. 29 

opposite forces T^ and T^. These two forces will have 
no effect as far as the state of rest or motion of the body 
on which the forces act is concerned. Combine these 

27 




28 APPLIED MECHANICS FOR ENGINEERS 

forces with Pj and P^^ respectively, obtaining the result- 
ants i^j and M^' i'hese resultants may be moved back 
to the point of intersection of their lines of action B, 
and resolved into components parallel to their original 
components. The two forces T^ and T^ then annul each 
other, and there are left the two forces Pj and P^ acting 
at B parallel to their original positions. The resultant then 
in case (a) is P = P^ + P^^ and in case (5) R== P^ — Py 

Let the line of action of the resultant cut the line A^A^ 
in (7, and let the distances from to A^ and A^ be re- 
spectively c?j and c?2* 

Then from similar triangles, in either case, 

BCP^ .BC_P^ 
a^ ij ^2 1 2 

By division, remembering that Tj = T2, 

d^ p; 

Hence, the resultant of two parallel forces acting in the 
same direction is equal to their sum^ is parallel to the forces^ 
and divides the line Joining their points of application in 
the inverse ratio of the forces. 

The resultant of two unequal parallel forces acting in 
opposite directions is equal to their difference^ acts hi the 
direction of the larger force^ and divides the lifie joining 
their points of application externally in the inverse ratio of 
the forces. 

26. The Moment of the Resultant of Two Parallel Forces. 
— Applying the theorem of Art. 24 for the moment of 



PARALLEL FORCES 29 

the resultant of concurrent forces to the forces of the 
preceding article, 

mom H = mom M^ + mom i^g 

= mom Pj + mom T^ + mom P^ + mom T^. 
But mom T^ + mom T^ = (Art. 24, Cor.). 

Therefore mom M = mom P^ + mom P^. 

Or, ^^Ag moment of the resultant of two parallel forces with 
respect to any line is equal to the algebraic sum of the mo- 
ments of the two forces ivith respect to that line. 

27. The Resultant of Any Number of Parallel Forces in 
Space. — By combining the resultant of two parallel forces 
with a third, and that resultant with another, and so on, 
the theorems of the two preceding articles may be ex- 
tended at once to any number of parallel forces : 

(1) The resultant of any number of parallel forces in 
space is equal to their algebraic sum and acts parallel to the 
forces. 

(2) The algebraic sum of the moments of any number of 
parallel forces in space with respect to any line is equal to 
the moment of their resultant with respect to that line. 

It folloAVS at once that if a set of parallel forces is in 
equilibrium^ the sum of the moments of all the forces with 
respect to any line is zero., since the resultant of the forces 
is zero, and hence the moment of the resultant is zero. 

If, conversely, the sum of the moments of a set of par- 
allel forces with respect to a line not parallel to the forces 
is zero, either the resultant of the forces must be zero, 
or else the line of action of the resultant must intersect 
the given line. If, however, the sum of the moments 



30 APPLIED MECHANICS FOR ENGINEERS 

of the forces about each of two lines not parallel to the 
forces which cannot intersect the same line parallel to 
the lines of action of the forces is zero, the resultant must 
be zero and the forces are in equilibrium. 

If the lines of action of the forces all lie in one plane, 
the most convenient line about which to take moments is 
perpendicular to the plane of the forces. 

The condition just stated for equilibrium in this case 
becomes : If the sum of the moments of a set of parallel 
forces in one plane tvith respect to each of three points of 




Fig. 30 

the plane not in the same straight line is zero, the forces are 
in equilibrium. 

Illustration. Forces of 7 lb., 5 lb., and 10 lb. act per- 
pendicular to the plane of the lines OA and OB, as shown 
in Fig. 30. The distances of the lines of action of the 
forces from OA and OB are respectively 1, 3, 4 units and 
2, 1, 5 units. To hnd where the line of action of the 
resultant cuts the plane OAB : 

Let a and b be the distances of the line of action of the 
resultant from OA and OB respectively. The result- 

^^'^^^ i2 = 7 + 5 + 10 = 22 1b., 

acting in the direction of the forces. 



PARALLEL FORCES 



31 



Taking moments about the line OA^ 

a.i2 = 1.7 + 3.5 + 4.10, 

or a = 2.82 units of space. 

Taking moments about OB, 

5. ^ = 2. 7 + 1-0+5. 10, 
or 6 = 3.14: units of space. 

Hence the resultant is equal to 22 lb., acts in the direc- 
tion of the forces, and at distances of 2.82 and 3.14 
units from OA and OB respectively. 




Fig. 31 

Problem 18. Two parallel forces, one of 20 lb. and one of 100 lb., 
have lines of action 24 in. apart. Find the resultant in magnitude, 
direction, and line of action : 

(1) When they are in the same direction. 

(2) When they are in opposite directions. 

Problem 19. A horizontal beam of length I is supported at its 
ends by two piers and loaded with a single load P at a distance of 

- from one end. Find the reactions of the piers against the beam. 

Problem 20. The locomotive shown in Fig. 31 is run upon a 
turntable whose length is 100 ft. Find the position of the engine 
so that the table will balance. 



32 



APPLIED MECHANICS FOR ENGINEERS 



Problem 21. Weights of 100, 50, and 120 lb. are placed at the ver- 
tices A , B, and C respectively of an equilateral triangle 4 ft. on a side. 
Find the distances bi the center of 
the forces from the lines AB and BC. 

Problem 22. On a square table 
weights of 70, 80, and 100 lb. are 
placed as shown in Fig. 32. Find the 
position of a fourth weight of 90 lb. 
which will balance the given weights 
with respect to the two lines AB and 
CD. 

Ans. f ft. from AB, 3f ft. from CD in quadrant BOD. 




Fig. 32 



28. Center of Parallel Forces. — In Art. 25 it was shown 
that the resultant of two parallel forces P^ and P^, acting 
at the points A^ and A^ respectively, divides the line A-^A^ 
in the inverse ratio of the forces. If, then, without 
changing the points of application of the parallel forces 



Ar 


C A, 






1 






t 

1 




1 


1 




1 

1 


Pi 


1 

1 


1 

1 




1 

it: 




1 
1 


1 
1 


Po 




1 


1 


" 




1 


1 






1 


1 






1 
1 
/ 


i__ 


• 




1 






1 
1 


R 




t . 










' 






-Pl+^2 



Fig. 33 



Fig. 34 



or their magnitudes their direction be changed, the point 
on ^1^2 through which the resultant passes is not changed, 
since it must still divide A^A^ in the ratio P^ : Py 

In the same way, if three parallel forces of fixed value 
Pj, -Pgi -^3 ^ct at three fixed points A^, A^, A^, and if C 



PARALLEL FORCES 



33 



is the point on ^1^2 through which the resultant of Pj 
and Pg passes, then the point C on O'A^ which divides 
C'Aq in the ratio P^:(^P^-\- P^ is a point through which 
the resultant of the three forces always passes no matter 
what their direction. The same reasoning applies to any 
number of parallel forces acting at definite points. For 
any set of parallel forces acting at definite points there is 
therefore a point through which the resultant always 
passes no matter what direction the parallel forces may 
take. This point is called the center of the parallel forces. 
29. Graphical Construction for the Center of Parallel 
Forces. — Let the parallel forces P^ and Pg act at A^ and 




A^ (Fig. 35). From A^ lay off A^B^ equal to Pg but 
in the opposite direction to Pg. From A^ lay off A^B^^ 
equal to P^ and in the same direction as Pj. The inter- 
section of B^B^ and A-^A^ is then the center of the two 
forces. For, letting A^C = d^ and OA^ = d^^ we have, by 
similar triangles A^B^C and A^Bfi, 

^ = 4A or ^ = ^ 

A* 



d. 



AB2 



d. 



"2 ^-^2^2 '^2 

Hence Ois the point found in Art. 25 to be the center of 
the forces. 



34 APPLIED MECHANICS FOR ENGINEERS 

If there are more than two parallel forces, the continued 
application of the construction will locate the center of 
the forces. 

Problem 23. Weights of 20, 45, and 70 lb. are suspended from 
points on a straight line at distances of 2, 5, and 7 ft. respectively 
from a point O. Find by graphical construction the center of the 
forces exerted by the weights. Check by the theory of moments. 

Problem 24. Locate graphically the center of the three forces in 
the illustration of Art. 27. (N.B. The forces may be assumed to 
act in the plane GAB without changing the position of the center.) 

30. In Regard to Signs. — Given a set of parallel forces, 
Pj, J*2^ -^3' ^tc-' acting at definite points. 

Let 01^ be a line in a plane perpendicular to the lines 
of action of the forces. For convenience assume the 
lines of action of the forces to be vertical. Let it be 
agreed that forces acting upward shall be positive and 
those acting downward negative. 

Let x-^^^ X2, % ••• be the perpendiculars from 01^ to the 

lines of the forces P^ P2' ^3^ •*•' 
X being counted positive if 
measured from the axis OY to- 
ward the right and negative if 
measured from OY toward the 
left. Then, when observed in 
the direction YO^ Px is the 

Fig. 36 p t^ • 1 

moment 01 F with respect to 
OY^ not only in numerical value but in sign. For if both 
P and X are positive, as P^ and x^^ or both negative, as Pg 
and a^g, the product is positive. Inspection of the figure 
shows that the tendency to rotation in both cases is counter- 
clockwise, which has been defined as positive. 




PARALLEL FORCES 



35 



If X is positive and P is negative, as x^ and P2' ^^ ^^ ^ 
is negative and P is positive, as x^ and P^, the product is 
negative, and an inspection of the figure shows that the 
tendency to rotation is clockwise, which has been defined 
as negative. 

The sum of the moments of the forces with respect to 

1^ is then PiX^ + ^2^2 + -^3^3 + -^4^4 + • ' ' • This sum 
is denoted by ^Px. 

31. Coordinates of the Center of a System of Parallel 
Forces. — Let Pp Pg, Pg, ••• be a set of parallel forces 
whose points of applica- 
tion are (x-^^ y^, 2^), (x^^ 

Vv ^2)' C^3' Vz^ %)' ••• ii^ 
rectangular coordinates. 

Denote the center of the 

forces by (2:, y, z). As 

was shown in Art. 28 the 

position of the center of 

the forces is independent 

of the direction in which 

they act. Assume then ^^^- ^^ 

that they act parallel to OZ, 

Taking moments about OY^ we have the moment of 

the resultant equal to the sum of the moments of the 

forces, or 

X{P^ + P2 + P3 + •••) = ^1^1 + ^^2^2 + ^3^3 +••• 

or x^P — ^Px. 

Taking moments about OX^ 

y^P=^Py, 




36 



APPLIED MECHANICS FOR ENGINEERS 



Assuming the forces to act parallel to OX and taking 
moments about OY, we obtain 



Therefore 



- _ 2 J>i/ 
2P ' 



y 



r_21>« 
^-2P^ 

Problem 25. Forces of 40, 65, and 70 lb. act in the same direction 
from the points (4, 3, 1), (5, - 2, 3), and (2, 3, 6), respectively. 
Find the resultant and the center of the forces. 

Problem 26. Three equal parallel forces act at the vertices of 
a triangle in the same direction ; prove that their resultant acts 
at the intersection of the medians of the triangle. Solve by applying 
the theorem of moments, and check by graphical construction. 

Problem 27. Four equal forces act in the same direction at the 
vertices of a regular tetrahedron. Find the center of the forces by 
taking moments. Also locate the center graphically. 

32. Arrang-ement of the Work for Computation. — If sev- 
eral parallel forces and their points of application are 
given, it is sometimes worth while to tabulate the work 
in some form such as the following : 



Forces 


Coordinates 


Moments 


X 


y 


Z 


Px 


Py 


Pz 


^1 


X, 


Vi 


^l 


P,x, 


PiVx 


p - 

1 y.^ 


P. 


X.2 


?/2 


^2 


-^2'^2 


^2.V2 


Pf-2 


Ps 


Xo 


?/3 


^3 


^3^3 


P^y-i 


P,h 


P. 


X, 


" //4 


-^4 


P,x, 


P^Va 


P,z, 
















Resultant 


X 


Jj 


~ 


^Px 


^Py 


^Pz 



PARALLEL FORCES 



37 



The values of quantities in the columns one, two, three, 
and four are given except in the last row. The values of 
the quantities in columns five, six, and seven are then 
computed and their sums and the sum of column one 
placed at the foot. The values of x, y^ and z are then 
easily computed. 

Problem 28. Using the above arrangement, find the coordinates 
of the center of the following parallel forces : P^ = 50 lb., P.^ = 100 lb., 
P3 = 300 lb., P4 = 10 lb., and P.. = - 400 lb., the points of applica- 
tion being respectively (2, 1, -5), (-1, -2, 4), (2, 1, -2), (-2, 1, 1), 
(1,1,1). Ans. (3, -4, -14). 

Problem 29. Parallel forces P^, Pg, P3, and P^ act at the 
corners of a rectangle 3 ft. by 2 ft. 
and perpendicular to its plane. Find 
the point of application of the result- 
ant, if Pj = 10 lb., P2 = 50 lb.. P., = 
100 lb., P4 = 200 lb., Pi and P2 be- 
ing 2 ft. apart, and Pg on same side 

as Pg. 



Problem 30. Eight parallel forces 
act at the corners of a one-inch cube. 
Find the point of application of the y- 
resultant force, if P-^ = 30 lb., Pg = 50 
lb., P3 = 10 lb., P4 = 20 lb., P5 = 100 

lb., P(i = o lb., P7 = 10 lb., Ps = 40 lb. The subscripts of the forces 
acting at the various vertices are shown in the figure (Fig. 38). 

Ans. (.283, .302, .415). 



Fig. 38 



CHAPTER IV 
CENTER OF GRAVITY 

33. Definition of the Center of Gravity. — The center of 
gravity of a body may be defined as the point of appli- 
cation of the resultant attraction of the earth for that 
body, and the center of gravity of several bodies consid- 
ered together, as the point of application of the resultant 
attraction of the earth for the bodies. The attention of 
the student is called to the fact that the forces acting 
upon the particles of a body, due to the attraction of the 
earth, are not parallel, but meet in the center of the earth. 
For all practical purposes, however, they are considered 
parallel. 

The center of gravity of many simple bodies can be 
found by inspection. For example, for a sphere of uni- 
form material it is evident that the line of action of the 
earth's attraction always passes through the center. For 
a rectangular block of uniform material the same is true. 

If a body be divided up into a number of parts whose 
centers of gravity are known, the whole weight of the 
body is the resultant of the known weights of the parts 
acting at known points, and hence the theorem of moments 
may be applied to determine the position of the center of 
gravity of the body; i.e. the equations 

-_2x^ ,7_2yP 5_2^ 

38 



CENTER OF GRAVITY 



39 



may be used where P^, P^^ etc., re^jresent the weights of 
the known parts, a;^, x^^ etc., the abscissas of the centers of 
gravity of these parts, etc. 

As an illustration consider the prob- 
lem of finding the center of gravity of 
the solid shown in Fig. 39. 

The figure represents a Z-iron of the 
same cross section throughout, and Pj, 
P^^ and Pg are the weights of the indi- 
vidual parts (considering the Z-iron as 
divided into three parts — two legs and 
the connecting vertical portion). If 
the w^eight of a cubic inch of iron = .26 
lb., Pi = .78 lb., ^2 = 2.08 lb., P3 = 1.04 
lb., and therefore i?=3.9 lb. The 
points of application of P^, P^^ and Pg are (— J^ — !» ^2)' 
Q, — ^, 5), and (2, — |-, |^), respectively, so that 




- _-78(-.i)+ -2.08(1) + 1.04(2) _ 



X = 



^ = 



3.9 



= .70 in.. 



.78(-i)4-2.Q8(-i:)+l.Q4(-i) ^ _ ^^ .^^^ 



3.9 

.78(V-)-f2.Q8(5) + l.Q4a) ^.^^ 
3.9 



in. 



This point x^ «/, z is, in this case, the center of gravity 
of tlie Z-iron. 

34. Centers of Gravity of Uniform Bodies and of Areas. — If 

Pj, P2, P3, etc., are the weights of the parts of a body or 
of several bodies, whose unit weights are, respectively, 
7^, 72' 73' ®tc., and volumes F^, V^^ Fg, etc., we may 



40 APPLIED MECHANTCH FOR ENGINEERS 

write JiV^ for P^, ry^V^ for P^, J^^s foi" -^3^ etc. The 
formulce for rr, «/, z then become 

~ ^ 7i ^1^1 + 72 ^^2^^2 + 73 ^3^3 + etc. ^ 27 Fa; 
7i ^1 + 72 ^^2 + 73 ^"3 + etc. l,yV ' 

- 27 FV - S7F2 

^ 27F 27F 

And if the bodies are all of the same material and so have 
the same heaviness, 7 is constant and may be taken outside 
the summation sign, Avhere it cancels out. This gives 

values for x^ «/, and z^ 

formulae exactly similar to those of Art. 33, where the P's 
are replaced by F's. 

If the bodies are thin plates of the same material, of 
constant thickness J, we may write for F^, F^, F^, etc. 
5Pj, hF^-, hF^^ etc., where the P's represent the areas of 
the faces of the plates. Making this substitution for the 
F's, x^ y^ z may be written 

- _ hF^x^ + ^^2^2 + ^-^3^3 + etc. _ ^Fx 
^~ 5Pi + bF^+bF^'+ etc. Yf' 

- _^Fy -^_^Fz 

y~ si' ^~ 2P' 

the 5, being a constant factor, cancels out. These formulge 
are taken as defining the '' center of gravity of an area " 
and are much used by engineers for finding the center of 
gravity of sections of angles, channels, T-sections, Z-sec- 
tions, etc. Usually, the a^y-plane is taken in the plane of 
the area and only the values of x and y are needed. 



CENTER OF GRAVITY 



41 



0.4 



Problem 31. 

Find the center 
of gravity of the 
channel section 
shown in Fig. 40. 

Problem 32. 

Find the center 

of gravity of the T-section 

shown in Fig. 41. 



-;^ 



-10'^- 

Fig. 40 



~X^ 




Fig. 41 



Fig. 42 



Problem 33. Find the center of gravity of the U -section shovrn 

in Fig. 42. Given the fact that the center of gravity of a semicircular 

4 r 
area is - — from the diameter. (See Prob. 41.) 

3 TT 

Problem 34. Find the position of the center of gravity of a 
trapezoidal area, the lengths of whose parallel sides are a^ and a^, re- 
rj p spectively, and the distance 

between them h. (See Fig. 
43.) 

Hint. Draw the diago- 
nal AB and call the tri- 
angle ACB, Fp and the 
triangle ABD, F^. 




Fig. 43 



Given, 

the center of gravity of a triangle is i the distance from the base to 
the vertex. (See Prob. 39.) Select .1 D as the a;-axis, then 



42 



APPLIED MECHANICS FOR ENGINEERS 



y 



_ ^ii/i±Z^ 



where y-^ = | h and y^= \ h. The center of gravity is seen to lie on a 

line joining the middle points of the parallel sides. 

Problem 35. A cylindrical piece of cast iron, whose height is 

6 in. and the radius of whose base is 2 in., has a cylindrical hole of 1 in. 
radius drilled in one end, the axis of which 
coincides with the axis of the cylinder. The 
hole was originally 3 in. deep, but has been 
filled with lead until it is only 1 in. deep. 
Find the center of gravity of the body, the unit 
weight of lead being 710 and of cast iron 450 
(Fig. 44). 

Problem 36. Find the center of gravity of 
a portion of a reinforced concrete beam. (See 
Fig. 45.) The beam is reinforced with three 
half -inch steel rods, centers 1 in. from the bottom of the beam and 1 

Z 




Fig. M 



^ 



0^:..C)1. 



-,S 



■X 



Fig. 45 



CENTER OF GRAVITY 



43 




m. 



H 



in. from the sides. The center of the middle rod is 4 in. from the 
sides. 

(y for steel = 490 lb. per cubic foot ; 
y for concrete = 125 lb. per cubic foot.) 

Note. It is seen that the thickness cancels out of the expression 
for the center of gravity, and might, therefore, have been neglected. 

35. Center of Gravity of a Body with a Portion Removed. — 

It is sometimes convenient in finding the center of gravity 
of a body to regard it as a larger body from which a por- 
tion has been removed. The weight of the given body 
can then be regarded as the resultant of the weight of the 
larger body acting at its center of gravity and a force 
equal to the weight of the portion 
removed acting at ' its center of 
gravity in the opposite direction. 

To illustrate this consider the 
cylinder of Fig. 46 of height jETand 
radius R from which a cylindrical 
portion of height h and radius r and 
having the same axis is removed 
from one end. 

If W is the weight of the whole 
cylinder, and w the weight of the portion removed, then 
the center of gravity of the remaining body is the center 
of the two forces W and w acting as shown in the figure. 
If 7 is the heaviness, 

W= yirE'^R, 
w = yirr^h. 

The distance y from the open end to the center of 
gravity of the body is therefore 



-R- 



-w 



Y 



\iM 



Fig. 46 



44 APPLIED MECHANICS FOR ENGINEERS 

- - 2^2 _ 1 722^2 _ ^2^2 

•^ ~ jttEW- yirr^h 2 E'^IT-r^h' 

In general the formula is, 

where TFg is the weight removed from the weight W^ 
and a:j and x^ are the abscissas of the centers of gravity 
of the weights W^ and TF2, respectively, before the removal 
of the weight W^- 

Problem 37. Through a circular disk 1 ft. in diameter a hole 
4" in diameter is bored with its axis distant 3" from the axis of the 
disk. Find the position of the center of gravity of the remainder. 

Problem 38. From two adjacent corners of a cube of edge 1 ft. 
cubes of 2-inch and 3-inchi edges are removed. Find the center of 
gravity of the remaining portion. 

36. Center of Gravity Determined by Integration. — In 
many cases of areas and solids the position of the center 
of gravity may be determined by integration. The method 
as applied to areas is as follows : 

Let F be any area in the 2:j/-plane. In order to deal 
with actual forces think of F as the face of a thin sheet 
of uniform thickness and uniform material and let 7 be 
the weight of the sheet per unit area in the surface F. 
Divide the area up into elements of area AF. (This may 
be done in a variety of ways, Fig. 47.) The weight of 
such an element is then yAF. The weight of the sheet 
may then be replaced by a set of parallel forces of magni- 
tude yAF acting at the centers of these elementary areas. 



CENTER OF GRAVITY 



45 



The centers of the elementary areas and the elementary 
areas themselves may be determined approximatel}', and 
if x' is the abscissa of the approximate center * of AF, 





then an approximate value for the abscissa of the center 
of the parallel forces is 

^'=%If- CArt.38) 

The limiting value of x' as the elementary areas are in- 
definitely decreased is defined as the abscissa, x^ of the 
center of gravity of the area F. Hence, 

- _ Lim ^x'yAF 
Lim l^yAF 

as AF approaches the limit zero. By a theorem of calculus 
this may be written 

Cx'ydF 

\dF ' 



x = 



/^ 



*The approximate center must of course be such that the approximate 
center and the true center of gravity of the element have the same 
limiting position as the elementary area is indefinitely decreased. If 
both dimensions of AF approach the limit zero as in {h) and (c), Fig. 47, 
the approximate center may be taken as any point of AF. 



46 



APPLIED MECHANICS FOR ENGINEERS 



or, since 7 is constant, 



Jx'dF 
fdF^ 



the integration being taken to inclnde the whole area F. 
Similarly 

Jy'dF 



y^ 



f 



dF 



In deriving the latter formula the forces would be 
thought of as acting parallel to the a;-axis. 

In the use of these formulae it must be kept in mind 
that x' and 7/' are the coordinates of the approximate 
center of the elementary area AF. 

Illustration, To find the center of gravity of the area 

bounded by the 

a;-axis, the curve 

i^^y}^ y = sin x^ and the 

line X ——' 

2 

Divide the area 
up into strips 
^ parallel to the 
?/-axis of width 
Ax. Then con- 
sidering any strip whose middle ordinate intersects the 
curve in the point (2:, ?/), the value of AF is yAx ap- 
proximately, the value of x^ is a:, and the value of y^ is 

I (Fig- 48). 




Fig. 48 



CENTER OF GRAVITY 



47 



Then 



- _ 

Jxydx I X sin xdx 
^ _ _o _»A 

J I ydx j 2 sin a:cZ2; 
Jo 



( — a; cos a: + sin x') 



cos a; 



1 '• 



J^ ydx - I sin^ a;c?a; 
2"^ _ 2-^0 

77 77 

J I ydx I sin a;6?a; 

*>'0 



(i ^ ~ i sin 2 a;) 



2 TT 



— COS a; 



8__7r 
T~" 8" 




Fig. 49 

Problem 39. Find the center of gravity of a triangle whose alti- 
tude is h and whose base is a. Take the origin at the vertex and 
draw the x-axis perpendicular to the base. (See Fig. 49.) 



48 



APPLIED MECHANICS FOR ENGINEERS 



ix'clF 
X = -J^ Here dF = aV/.r, and from similar triangles 

^4F 



a' — ~ X, so that dF = - xdx. 
h A ' 



and 



sc 



h 

a rh 
Ji 



3Jo 



\ x\lx 
> 

Cxdx~'^-X 
Jo 2 Jo 



I- 



The center of gravity is | the distance from the vertex to the base, 
and since the median is a line of symmetry, it is a point on the 
median. It is, in fact, the point where the medians of the triangle 
intersect. 

B 




Fig. 50 



Problem 40. Find the center of gravity of a parabolic area 
shown in Fig. 50, the equation of the parabola being if = 2 px. 
Here dF = ydx, so that 

i xydx ■\/2p i '^x^dx -5 x^ \ g 

("a ra 1 -3 Ho 5 

\ ydx v-/>\ x'^dx I ^^ L 

It is left as a problem for the student to show that y = \h. 

Problem 41. Find the center of gravity of a sector of a flat ring 
outside radius R^ and inside radius R^. (See Fig. 51.) Let the 



CENTER OF GRAVITY 



49 



angle of the sector be 2 6. Take the origin at the center and let the 
a:-axis bisect the angle 2 0. 

Here dF = pdpda, and x = p cos a, so that 

I i cos ada • p^dp 



X = 




Fig. 51 
Integrating the numerator first, 

\ cos ada r J,/p = ^1 .^^^ cos «cZ« - ^^ , ^'^ (2 sin (9). 

Integrating the denominator, 



_ 2 JJi3 - JBs^ sill 



Therefore, ^ 3 «,. _ jj,. e 

If i?2 = 0, the sector becomes the sector of a circle, and x becomes 

_ 2^ sin^ 

If the sector is a semicircle, that is, if 2 6 = tt, then, since = -, 

3 IT 



50 



APPLIED MECHANICS FOR ENGINEERS 



Problem 42. Find the center of gravity of a semi-ellipse (Fig. 52) 
whose equation is 



.r^ . // 






therefore 



(IF = 2 ydr = 2 - V'a^ - x^ - dx, 
a 




Fig. 52 



hra 

2 ~\^ ^\/(jfi — 'xP- dx 



-lia^-x'y]l 



X ■= 



]) fa If X]" 

2- I Va'^ — x'^dx H xVa^ — a:^ + rt2sin-i~ I 



3 _4a 



O^ TT 3 IT 

2 *2 



CENTER OF GRAVITY 



51 



Problem 43. Find the center of gravity of the area between the 
parabola, the ^/-axis, and the line AB in Problem 
40. 

Problem 44. A quadrant of a circle is taken 
from a square whose sides equal the radius of the cir- 
cle. (See Fig. 53.) Find the center 
of gravity of the remaining area. 

Problem 45. Suppose that the 
corners A and C of the angle iron 

in Fig. 54 are cut to the arc of a circle of j\ in. radius 
and the angle at B is filled to the arc of a circle f in. 
radius; what would be the change in Ic and ^? 



Y 




Fig. 53 



J3 



C 



T 



X 



Fig. 54 



37. Center of Gravity of a Solid. 

— By dividing a solid into in- 
finitesimal parts and taking the 
moments of the weights of these 
parts with respect to the three 

coordinate axes, the coordinates of the center of gravity 

of the solid may be found. 

As an illustration, suppose it is desired to obtain the 

center of gravity of a right circular cone of altitude h and 

radius of base r. 

Take the a>axis as 

the axis of the cone 

with the vertex at 

the origin. (See 

Fig. 55.) It is 

evident, that ^ = 

and 2 = 0, so that 

it is only necessary 

to find X. The Fig. 55 




52 APPLIED MECHANICS FOR ENGINEERS 

volume, dv, cut from the cone by two parallel planes, per- 
pendicular to OX and separated by a distance dx, is ir^^dx, 
and the weight of this dv is ^iry^dx — dP. Therefore 



X = 



I x' dP i x^iry^d:* 



J dP I ^iry^dx 



But from similar triangles y : x::r :h or y = -x. This 
gives 



y2 /»A 



rr*' 



A2 



>2 /'A /v^" 



3, 

-= -rh. 

h 4 



The expressions for ^, ^, and 2, involving c?P, ma}^ be 
changed to similar ones involving dv^ and these become 
for homogeneous bodies, since dP = ^dv^ 

ix'dv _ iy'dv _ iz'dv 
^ -_ ^1 , y = ^ , z = ^ 

j 6??; i ds I dv 

The center of gravity of thin homogeneous wires of con- 
stant cross section may be found by replacing the dv in 
the above formulae by ads^ where a is the constant area 
of cross section and ds is a distance along the curve. The 
formulae then become 



j x'ds I y' ds { z' 



ds 



a?=^ — , y = 



fds' fds f 



ds 



Problem 46. Find the center of gravity of a hemisphere, the 
radius of the sphere being r. Let the equation of the generating 
circle of the surface be x^ -\- y^ = r^. Divide the hemisphere into 



CENTER OF GRAVITY 



53 



slices of thickness 


dx by planes parallel to the plane face of the 


hemisphere. 




Then 


(x'dP 

X — —^ , where dP = y-nyHx = y7r(r'^ — x'^)dx. 

J4P 


Show that 


'=h 




Fig. 56 



Problem 47. Find the center of gravity of a portion of circular 
wire (Fig. 56) of length L and whose chord = 2 b. Take the center 
of the circular arc as oricrin and let the x'-axis bisect L. Then 



- J 



xds 







X 


^ds 






But 


.r^ 


+ .y2 


= RK 






.*. 2 


xdx + 


.'. ds 


= 0. 








\x/ 






= y/dx^ + dy'^ = 


■dy 








= —dy. 

X 










.*. T 


R\dy 


_ radius x 


chord 



L 



arc 



54 



APPLIED MECHANICS FOR ENGINEERS 



For a semicircular wire 



X — 



Diameter 



TT 



Problem 48. Find the center of gravity of a paraboloid of revo- 
lution. If the equation of the generating curve is y^ = 2px, and the 
greatest value of x is a, show that 

3 

Suggestion. Use the same method as that used for the right 
circular cone. 

Problem 49. (a) Show that the center of gravity of the circular 
sector A OB (Fig. 57) of angle 2 a and chord 2 d is given by 

_ _ 2 d _2 radius x chord 
3 a 3 arc 




Fig. 57 



(b) From this result and the known position of the center of 

gravity of the triangle, prove that the center of gravity of the segment 

ADB is given by 

-,^2d^ 

^ ~SF 



where F is the area of the segment. 



CENTER OF GRAVITY 55 

Suggestion. In (a) use polar coordinates. The element of area 
is pdOdp and 

("Cp'' cos ec/Odp 



X = 



i^i^'^'f 



Problem 50. Show that the center of gravity of any pyramid is 
in a plane parallel to the base which cuts the altitude at | the dis- 
tance from vertex to base. 

Problem 51. Show that the center of gravity of a hemispherical 
surface bisects the radius perpendicular to the plane of the base of 
the surface. 

Problem 52. Show that the center of gravity of a spherical zone 
is midway between the planes of the bases of the zone. 

Problem 53. Show that the center of gravity of the surface of a 
right circular cone is at | the distance from the vertex to the base. 

Problem 54. Show that the center of gravity of the frustum of a 
right circular cone, the radii of the bases being R and r and the alti- 
tude H, is distant 

H(R^ + 2Rr+S r^) 
4:(R'^ + Rr + r2) 

from the base of radius R. Find also the distance of the center of 
gravity from the base of radius r. 

Problem 55. A casting is in the form of a hollow cylinder with 
one end closed. The thickness of the end is | in., the length of the 
casting is 12 in., and the radii of the inner and outer surfaces of 
the cylinder are 5^ and 6 in. Find the position of the center of grav- 
ity of the casting. 

Problem 56. If the above casting is filled with material | as 
heavy as the material of the casting, find the position of the center of 
gravity. 

Problem 57. A hemispherical shell of inner and outer radii r 
and R rests upon a hollow cylinder of the same material of height H 



BQ 



APPLIED MECHANICS FOR ENGINEERS 



and inner and outer radii ?• and R. Find the distance of the center 
of gravity from the base of the cylinder. 

Problem 58. From a riglit circular cone of height 20 in. and 
radius of base 10 in. a cylinder of height 10 in. and radius of 
base 3 in. is removed, the base and axis of the cylinder being in 
the base and axis of the cone. Find the distance from the base of 
the cone to the center of gravity of the part remaining. 

Problem 59. If the cylindrical portion of the preceding problem 
is filled with material n times as heavy as the material of the cone, 
find the distance from the base to the center of gravity. 

38. Center of Gravity of Counterbalance of Locomotive Drive 
Wheel. — In Fig. 58 the drive wheel is indicated by the 
circle and the counterbalance by the portion inclosed by 
the heavy lines, the point is the center of the wheel, 
and a is the angle subtended by the counterbalance. The 




Fig. 58 

point 0' is the center of the circle forming the inner 
boundary of the counterbalance, and /3 is the angle sub- 
tended by the counterbalance at this point. Let F^ repre- 
sent the area of the segment of radius r and F^ the area 
of the segment of radius r^ Also let x-^^ represent the 
distance of the center of gravity of F^ from 0, and x^' the 



CENTER OF GRAVITY 57 

distance of the center of gravity of F^ from 0' . Then, 
from Problem 49, 

X. = 77— =r and .r^' — — — . 

But a^gi the distance of the center of gravity of Fc^^ from (9, 

= Xo' — 00 = —-—- — r. cos ^ — r cos - . 

^ 3 i<^2 V 2 27 

Then the distance of the center of gravity of the counter- 
balance from is 



X = 



^1-^2 



_^2 



/3 a 

r-, cos ^ — r cos - 



Q 






where F. = ar cos -, 

^ 2 2 

and #„ = ^— J — ar-t cos — . 

Problem 60. If r = 3 ft, r^ = 10 ft., and a =120^, find the posi- 
tion of the center of gravity of the counterbalance of Art. 38. 

39. Graphical Method of Finding the Center of a Set of 
Parallel Forces in One Plane. — Let P^, P^, Pg, P^ be a set 

of parallel forces acting at the points A^, A^^ ^g, A^^ 
respectively, in one plane. Assume the forces to be acting 
in the direction shown in Fig. 59. Let one of the forces, 
as Pj, be moved to any point in its line of action, as JB^ 
and resolved into two components in any two directions, 
as S^^ J and S^^- Extend the line of one of these compo- 



58 



APPLIED MECHANICS FOR ENGINEERS 



nents, aS'^^ g' ^^ ^^^ ^^^^ li^^ *^f ^ second force, Pg' ^^^^ 
resolve the second force into two components, one of 
which is equal and opposite to the component, S^^ ^, of 
the first force on that line. Extend the line of the other 
component, S^^ 3, of the second force, to cut the third 




Fig. 59 

force and resolve the third force into components as in 
the case of the second force. Proceeding in this way 
until the last force is reached there remain then only two 
unbalanced components of the original forces, one of the 
components, S^^ -^^ of the first force, and one, S^^ 5, of the 
last force. The resultant of the original forces is there- 
fore the resultant of these two unbalanced components. 
By Art. 27 the resultant of the parallel forces is parallel 
to the forces. Hence the resultant, P^^ acts in a line, 
parallel to the original forces, through the intersection, 
^g, of the unbalanced components, aS^^^ j and S^^ g. 



CENTER OF GRAVITY 59 

The actual work of locating the line of action of the 
resultant may be shortened by the following considera- 
tion : the triangles of forces at the vertices B^^ B^^ B^^ 
B^ may be placed with equal sides coinciding, as in 
Fig. 59 (5). 

The sides representing the forces P^, P^^ Pg, P^ then 
fall in succession on a straight line, the beginning of any 
force falling at the end of the preceding one. The lines 
representing the components S-^^^^ S^^ g, etc., all meet at a 
common point 0. The rays from are then parallel 
respectively to the sides of the polygon B^B^B^B^B^. 
This enables one to draw the polygon B^B<^B^B^B^ 
without constructing the separate triangles B^B^Up etc. 
Since S-^^ ^ ^^^ ^5, i ^^^ arbitrary directions given them, 
the point may be chosen arbitrarily, and the polygon 
B^B^B^B^B^ constructed by drawing its sides parallel 
to the corresponding rays from 0. To sum up, the 
method is as follows : 

Assume a direction in which the forces act. Lay off in 
succession the forces on a line parallel to their lines of 
action. Join their points of intersection with an arbitrary 
point 0, Fig. 60 (h). Construct a polygon with vertices 
on the lines of action of the forces and with sides parallel 
to the corresponding rays from 0. The intersection of 
the sides of this polygon, B^ in Fig. 60 (a), drawn par- 
allel to the rays from to the beginning and end of 
the line of forces, is a point through which the resultant 
passes. This gives the line of the resultant for the as- 
sumed direction of action of the forces. The center of 
the forces is therefore on this line. 



60 



APPLIED MECHANICS FOR ENGINEERS 



In the same way another line of action for another as- 
sumed direction may be determined. The intersection of 
these two lines is the center of the given forces (Art. 28). 




Fig. 60 

In doing this work the forces may be laid off in suc- 
cession in any order. It is only necessary to make the 
sides of the polygon B^B^B^ ... correspond to the 
rays from to the line of forces; i.e. the side of the 
polygon connecting any two forces must be parallel to 
the ray from to the intersection of the same two forces. 

The polygon B^B^B^ ... is called the equilibrium 
polygon. A more general discussion will be given in 
the chapter on non-concurrent forces in one plane. 

Problem 61. Parallel forces of 8, 12, 16, and 20 lb. act at 
points (0, 0), (1, 3), (3, I), and (5, 4) respectively. Find graphically 
the line parallel to the ?/-axis in which the center of the forces lies. 
Make the construction twice, laying off the forces in different orders. 
Check by the theorem of moments. 

Problem 62. Find graphically the distance from the ar-axis to 
the center of the forces of the preceding problem. 

Problem 63. AVeights of 1200, 1600, 3000, and 2000 lb. act on a 
beam at distances of 0, 5, 8, and 12 ft. respeetively from one end. 



CENTER OF GRAVITY 



61 



Find graphically where their resultant cuts the beam. Check by 
moments. 

Problem 64. Locate graphically the center of gravity of the area 
of a quadrant of a circle. 

Suggestion. Divide the area up into a number, say eight, of 
strips of equal width (Fig. 61). The weights of the strips are then 
approximately proportional to the ordinates at 
their middle points and act in the lines of these 
ordinates. Hence use these ordinates, or any 
proportional parts of them, as forces and pro- 
ceed as in finding the center of a set of parallel 
forces. 




Fig. (51 



Problem 65. Locate graphically the center 
of gravity of the arc of a quadrant of a circle. 

Problem 66. Show how to find graphically, 
approximately, the center of gravity of any area in one plane. 
Apply the method to finding the center of gravity of a given area. 
Check by cutting the area out of cardboard and balancing. 

Problem 67. Find graphically, approximately, the position of 
the center of gravity of a hemisphere. 

Suggestion. The hemisphere may be divided up into a number 
of parts by parallel planes equally spaced. The w^eights of these 
parts will then be approximately proportional to the squares of the 
ordinates at the middle of the generating strips of area. Lines pro- 
portional to the squares of these ordinates may be constructed as 
shown in Fig. 62 (b) . These lines may then be used as forces and 
the graphical solution obtained as in the preceding problems. 

Problem 68, Find graphically the position of the center of gravity 
of a solid of revolution obtained by revolving a parabola about its axis. 

Problem 69. By the method of Problem 67 find the position of 
the center of gravity of the frustum of a cone. Compare the result 
with that obtained from the answer to Problem 54 by giving particular 
values to it, r, and H. 



62 



APPLIED MECHANICS FOR ENGINEERS 





(a) 



Fig. 62 



40. Simpson's Rule. — When the algebraic equation of a 
curve is known, it is expressed as y =f(^x)^ and the area 
between the curve and either axis is usually determined 
by integration. In Fig. 63 the area ABCD is expressed 
by the integral 

ydx = 1 f(x)dx, 

a *^ a 

r 




Fig. 63 



CENTER OF GRAVITY 



63 



when the curve represented by y =f(x) is continuous 
between A and B. 

In many engineering problems the curve is such that 
its equation is not known, so that approximate methods of 







/-. /I 


1 — ] 








n 


G 1^ 


1 " 














A 


n 


^ 




















^ 


Vo 


y^ 


y. 


2/3 

E 


2/4 


y, 

F 


y. 


2/7 


y. 


y. 


2/io 






7. 






' 
















u 















Fig. 64 



F 



G 




H 
Fig. 65 



D 



obtaining the areas under the curve must be resorted to. 
One of these methods of approximation is known as 
Simpson s Rule. Suppose the curve in question is the 
curve AB (Fig. 64) and it is desired 
to find the area between the portion 
AB and tlie a;-axis. Divide the length 
h — a into an even number, /^, of equal 
parts (here 7i=10). Consider the 
portion CDEF and imagine it magni- 
fied as shown in Fig. 65. Pass a 
parabolic arc through the points C^ 
G^ i), the axis of the parabola being 
parallel to the ?/-axis ; then the area 
CDEF is approximated by the area 
of the parabolic segment OGDI plus 
the area of the trapezoid CDEF. 



F 



64 APPLIED MECHANICS FOR ENGINEERS 

.-. area CaDi:F = \iy^ + y.^EF + | [2/3 - \ {y^ +3/4)]^'^^, 
since the area of the parabolic segment is | the area of the 

circumscribing parallelogram. Since EH = = Aa;, 

this area may be written 

In a similar way the next two strips to the right will have 
an area, -^(^4 + 4 ^/^-f-^/g), and the next two strips, an 

o 

\nr. 

area, -^(^g + 4 ?/y + ?/g), and so on. Adding all these so 

o 

as to get the total area under the portion of the curve AB^ 
we get 

total area = ^^ [y^ + 4(?/i + 2/3 + ^/^ + y^ + y^) 

+ 2(2/2 + ^4 + ^6 + J/s) + ^10]' 
or in general for n divisions, 

total area = -^ [?/o + 4(^i + ^^ + ^ 4- ... ?/„_i) 

+ 2(^/2 + ^4 + «/6 + ••• 2/»-2) + ^n]. 

and this is Simpson's formula for determining approxi- 
mately the area under a curve. It is easy to see that 
the smaller Aa:, the closer the approximation will be. 

41. Application of Simpson's Rule. — Simpson's Rule may 
be made use of in determining approximately not only 
areas, but volumes and moments. For if a volume be 
divided up by planes perpendicular to the a:-axis, distant 
Aa; apart, into elements of volume A?;, then dv = Adx, 
where A is the area of the cross section at any value of x^ 



CENTER OF GEAVITT 



Qb 



and the volume from x = a to x=h may be expressed as 

JAdx. Hence the volume may be evaluated by Simp- 
a 

son's formula, the y of the formula being replaced by A. 
The sum of the moments of these elements of volume 

with respect to the ?/-axis is j xdv or I xAdx and may be 

evaluated by Simpson's formula, replacing y by xA. On 
account of its use in adding moments Simpson's formula 



^0 


^•li 


A„ 


^u 


.4 


^ X\ y\ y^\ 


y 




1 
1 
1 

1 
1 
1 
1 

1 
1 


1 
1 




-;^ 




1 




J- -" "^ 


^ 


y^ 


1 




^'^'' y^ -^^^^ 




• "*- . 


-~— 4— - 


^ 


/j^^^**""^ 




^ 


y 












^ ^^^^^^^^ 







Fig. 66 

may be employed in finding the center of gravity of areas 
or volumes bounded by lines or surfaces whose equations 
are not known. Suppose, for example, it is desired to 
know the volume and position of the center of gravity 
of a coal bunker of a ship as shown in Fig. ^^. The 
bunker is 80 ft. long and the areas in square feet are as 
follows: A^ = 400, A^ = 700, A^ = 650, A^ = 600, A^ = 400. 
The distance between the successive areas is 20 ft. 
Applying Simpson's formula for volume, 
80 



volume = 



(8) (4) 



lA, + 4CA, + A,-) + 2A, + A,]. 



P 



6Q 



APPLIED MECHANICS FOR ENGINEERS 



Summing the moments, we obtain 
80 



^vx 



(3) (4) 



[AqXq + 4c(A^x^ + A^x^-) + 2^122^2 + A^J' 



where Xq = 0, 2-^ = 20, x^ = 40, x^ = 60, x^ = 80. The po- 
sition of tlie center of gravity from the fore end can now 
be obtained from the relation 



'liVX 

x = 



An approximate value of x may also be obtained from 
the equation 

^1 + ^2+^3 + ^4 

where v^ = 1(Aq + -4^)20, v^ = ^(A^ + A^}20, etc., 

and x\ = 10, x'^ = 30, etc. 

Problem 70. A reservoir with five- 
foot contour lines is shown in Fig. 67. 
Find the volume of water and the distance 
of the center of gravity from the surface 
of the water, if the areas of the contour 
lines are as follows: ^o = 0, ^^ = 100 sq. 
ft., ^2 = 200 sq. ft., ^3 = 500 sq. ft., A^ = 
600 sq. ft., A^ = 1000 sq. ft., Aq = 1500 sq. 
ft., ^- = 2000 sq. ft., ^8 = 2500 sq. ft. 
Making substitutions in the Simpson's 
formula, it becomes, for the volume, 




volume = 



[^0+4(^1 + A^ + A,+ A,)+2(A,+ A^-\-Aq)+A^-]. 



(3) (8) 

Summing the moments by Simpson's formula, we have 
40 



'^vx = 



(3) (8) 



+ 2(^422:2 + ^4^4 + ^c^<;) + ^s^s]^ 



CENTER OF GRAVITY 



67 



where a;o = ft., x^ = 5 ft., x^ = 10 ft., etc. Then 



- _ '^vx 

V 

Both numerator and denominator are computed by Simpson's formula. 

Compute X by means of the equa- 
tion, 

- _ v-jX'^ + Vqx'^ + v^x'^ + ••• 

^1 + ^'2 + *'3 + ••• 

and compare with the previous re- 
sult. 

Problem 71. Compute x for the 
parabolic area of Fig. 50, by using 
Simpson's Kule, and compare the 
result with that obtained by integra- 
tion. 

Problem 72. By Simpson's ^^^- ^^ 

Rule find the area and center of gravity of the rail section shown in 
Fig. 68. 

Beginning at the top, the horizontal distances are as follows : 




n,, = 4" 


wg = 1.24" 


u^= 1.0" 


Wii = 4.08" 


u^ = 1.18" 


u^ = 1.24 


Wjo = 4.24" 


u, = 1.0" 


u, = 2.23 


u^ = 2.5" 


u, = 1.0" 


u^ = 5.5" 
n, = 6" 



Problem 73. Find the center of gravity of the deck beam section 
shown in Fig. 69. Use the equation 

- _ F^x\ + F^x'^ + F.,x'^ + etc. 
i^i -}- ^2 + i^3 + etc. ' 

dividing the area of the section into convenient areas. Check the 
result thus obtained with that obtained by balancing a stiff paper 
model over a knife edge. 



68 



APPLIED MECHANICS FOR ENGINEERS 



-1.91- 



42. Durand's Rule. — A method of find- 
/ N ing the area of irregular areas was pub- 
^ '^ lished by Professor Duraiid in the Un- 
*' gineering News^ Jan. 18, 189-4. The rule 
states that the total area of an irregular 
curve equals 

-^ (0.4 3/q + 1.1 z/^ + ^2 + ^3 + 3/4 + •• • 

Jr 1.1 y^_,+OAy,) 

where a, 5, n, and the ^'s have the same 
meaning as in Simpson's Rule. The 
number of divisions may be even 
or odd. The student is advised 
to make use of this rule as well as 
Simpson's Rule and compare the 

Fig. 69 results. 




43. Theorems of Pappus and Guldinus. — Let -F be a plane 
area and OX a line in its plane not cutting the area F 
(Fig. 70). Let the 
plane containing the 
area F be revolved 
through the angle a 
about the line OX 
as an axis. Let dF 
be an element of the 
area at a distance y 

from OX. Then the _ _ 

element of volume ^^^- "^^ 

generated by dF is approximately dv = yadF, and the 



y^ ^^ 


Y^ 


dF 


\ 


r ,/^ /^ 


y^ 


\ 


\ 


\^\^ 








\ 


\ \ 




. \ ^ 


\ 


1 

y 




^ 


\ 




\ 


'- — ^ 



CENTER OF GRAVITY 69 

total volume generated by the area F is 

V=Ja7/dF 

= ai ydF. 

But the distance of the center of gravity of the area F 

CydF 

from OJTis given by y =^^—— — • 

F 



■f> 



ydF = Fy, and 

V=Fay. 

This may be stated as a general principle as follows : 
The volume of any solid of revolution is equal to the area 

of the generating figure times the distance its center of 

gravity moves. 

Problem 74. Find the volume of a sphere, radius r, by the above 
method, assuming it to be generated by a semicircular area revolving 
about a diameter. 

Problem 75. Assuming the volume of the sphere known, find the 
center of gravity of the generating semicircular area. 

Problem 76. Find the volume of a right circular cone, assuming 
that the generating triangle has a base r and altitude h. 

Problem 77. Assuming the volume of the cone known, find the 
center of gravity of the generating triangle. 

Problem 78. The parabolic area of Problem 40 revolves about 
the ic-axis ; find the volume of the resulting solid. 

Problem 79. Find the volume of an anchor ring, if the radius of 
the generating figure is a and the distance of its center from the axis 
of revolution is r. 



70 



APPLIED MECHANICS FOR ENGINEERS 




Fig. 71 



Let the curve AB (Fig. 
71), of length Z, be the gen- 
erating curve of a surface 
of revolution. The area 
of the surface generated by 
ds when revolved through 
the angle a is dF= ayds^ 
and the area of the whole 

surface is i<^= a j yds. The center of gravity of this curve 

AB is given by the expression 

I yds 

.'. I yds = ly^ and 
F = lay. 

This may be stated as follows : The area of any surface of 
revolution is equal to the length of the generating curve times 
the distance its center of gravity moves. 

Problem 80. Find the surface of a sphere, radius r, assuming the 
generating line to be a semicircular arc. 

Problem 81. Find the center of gravity of a quadrant of a circu- 
lar wire, radius of the circle r. Use results obtained above. 

Problem 82. Find the surface of the paraboloid in Problem 78. 



\jyd,. 



CHAPTER V 



COUPLES 




Fig. 72 



44. Definitions. — Two numerically equal forces acting 

in parallel lines in opposite directions form a couple. 

The distance between the forces 

is called the arm of the couple. 

The product of one of the forces 

and the arm, with the proper 

sign prefixed, is called the 

moment of the couple. The sign 

is plus if the forces tend to 

produce counter-clockwise rotation and negative if they 

tend to produce clockwise rotation. If P is one of the 

forces, d the distance between 
them, and M the moment of the 
couple, then M= ± Pd. 

The unit moment is that due 
to unit forces at unit distance 
apart. If P = 1 lb. and c? = 1 
ft., the moment is called 1 lb. -ft. 

45. Moment of a Couple about 
^^^- "^'^ Any Point in its Plane. — Let be 

any point in the plane of the couple of Fig. 73 and let a 
be the distance from to one of the forces, P', as shown. 

71 




72 APPLIED MECHANICS FOR ENGINEERS 

(The forces are marked P and P' to distinguish one from 
the other; P = P'.) Then for the point the sum of the 
moments of the forces P and P^ with respect to the point is 

P(a + (^)-P'a, or P^; 
for the position 0-^ the sum of the moments is 

P(d-a)-\-F'a, ov Fd ; 
for the position 0^ the sum of the moments is 

P'a - P(a - d), or Pd. 

Hence, the sum of the moments of the forces forming a 
couple with respect to any point in the plane of the couple is 
equal to the moment of the couple. 

46. Combination of Couples in the Same Plane. — Let two 

couples of moments M^ and M^ act in the same plane. 

The forces forming the 

^ couples may always be 

combined in pairs, one 

force from each couple, 

into two numerically 

equal, parallel forces 

opposite in direction, 

R and R' (Fig. 74). 

^^^- ^^ These resultant forces 

therefore form a couple. (If they fall in the same line, 

the moment of the couple is zero.) 

Let MhQ the moment of this resultant couple. If any 
point in the plane be chosen, then with respect to this 
point, 

M= mom R + mom R' . 




COUPLES 



73 



But mom R = mom P^ + mom P^ 

and mom W = mom P-^ H- mom P^ . (Art. 22.) 

Therefore, adding, 
itf = (mom Pj + mom P^') + (mom P^ + mom PgO 

Therefore, ^^o couples in the same plane are equivalent to 
a single couple in their plane whose moment is equal to the 
algebraic sum of the moments of the two given couples. 

It follows at once that any number of couples in the 
same plane are equivalent to a single couple in that plane 
whose moment is equal to the algebraic sum of the 
moments of the given couples. 

47. Equivalent Couples, (a) In the same plane. — Let 
the forces P^ and P^' form a couple whose moment is 
M^ = P^dy Let any two 
parallel lines AB and CD dis- 
tant d^ apart cut the lines of 
action of P^ and Pj' in A 
and C respectively. Move 



the forces P^ a 



nd P/ to A 



and and at these points 
put in two equal and oppo- 
site forces, S and S', in the 

line AC, of such value that 

« 

the resultant of Pj and S 
falls in the line AB. The 
resultants, P^ and P^', of P^ and S and of Pj' and S' re- 
spectively, are then numerically equal, parallel, and oppo- 
site in direction. They therefore form a couple. 




Fig. 75 



74 



APPLIED MECHANICS FOR ENGINEERS 



By similar triangles, Fig. 75, 

Therefore -f*2^2 — P^d^. 

Hence, a couple may he replaced hy any other couple in 
its plane provided only that the moment of the second couple 
is equal to the moment of the first. 

(5) In parallel planes. — Let P and P' be the forces of 
a couple in the plane MN. From two points A and A' in 
their lines of action draw parallel lines to cut the plane 




Fig. 76 

M'JSJ'', parallel to MJST, in the points B and B' . At B and 
B' put in pairs of equal and opposite forces P and P' 
parallel to the forces P and P' in the plane MN. 
The two forces P and P at ^ and B^ may be combined 
into a single force 2P acting at the middle of the 
line AB'. Likewise P' and P' acting at A' and B 



COUPLES 



75 



may be replaced by the force 2P' acting at the middle 
of A'B. But ABA' B' is a parallelogram, and the 
diagonals bisect each other. Therefore the forces 2 P 
and 2 P' act at the same point, and, since they are equal 
and opposite, annul each other. There are left then the 
forces P at ^ and P' at B\ forming a couple in the plane 
M'N' ^ equal in moment to the original couple. Hence a 
couple may be moved to any plane parallel to the plane in 
which it acts. 

To sum up (joi) and (K) : A couple acting on any body 
may he replaced hy any other couple of the same moment 
acting anywhere in the plane of the couple or in any par- 
allel plane. 



48. Moment of a Couple with Respect to Any Line. Defi- 
nition. — The moment of a couple with respect to any line 
is the sum of the 
moments of the 
forces forming that 
couple with respect 
to that line. 

Let P and P' form 
a couple in the plane 
HN whose moment 
is M= Pd^ and let 
AB be any line mak- 
ing the angle a with 
the normal to UN. Pass a plane RT perpendicular to AB 
to intersect the plane UN in HV. The couple may be 
moved in its plane until the forces are parallel to the line B. V 




Fig. 77 



76 



APPLIED MECHANICS FOR ENGINEERS 



(Art, 47) without changing the moment of the couple with 
respect to AB. For (Fig. 75) with respect to any line, 

mom P^ = mom P^ -f- mom S 
and mom P^' — '^^om P^ + mom S' . (Art. 24.) 

Adding, 

mom P^ + mom P^ = mom P^ + mom P^'^ 
since mom S + mom aS'' = 0. 

The projections of the forces P and P' on the plane ITT 
are equal to the forces themselves, and the projection of 
the distance, d, between them is d cos a. The moment of 
the couple with respect to AB is therefore Pd cos a, or 
Mcos a (Arts. 23 and 45). 

49. Combinations of Couples in Intersecting Planes. — Let 

the couples of moments Jf^ and M^ (Fig. 78) be replaced 

in their planes by 
equivalent couples 
having a common 
arm, a, coinciding 
with the line of 
intersection of the 
planes (Art. 47). 
The forces forming 
the couples may then 
be combined into a 
pair of numerically 
equal, parallel, and 
opposite forces B, 

and B' which form a couple in a plane containing the line 

of intersection of the two given planes. 



i^E7 




Fig. 78 



COUPLES 



77 



If a is the angle between the planes, measured between 
Pj and Pg (Fig- '^^)' ^^^ ^is the moment of the resultant 
couple, then 

M= aR = a VPi2 + P^s + 2 F^F^ cos « 
= V(aPi)2 + (aP2)2+ 2 aP^aPg cos a 

' = -y/M^ ■\-M^ + 2 M^M^ cos a. 

The angle, ^, which the plane of the resultant couple 
makes with the plane of the couple of moment M-^^ meas- 
ured from P-^ to P, is given by 



or 



tan 6 = 



tan 6 = 



Pg^sin a 



Pj + Pg cos a 
M^ sin a 



M^ + M^^ cos a 

Conversely, the couple of moment itf may be regarded as re- 
solved into the component couples of moments M-^ and M<y^. 

50. Vector Representation of Couples. — All of the prop- 
erties of couples derived in the preceding articles may be 
represented geometrically by 
regarding the couple as a 
vector quantity as follows ; 
A couple may be represented 
by a vector perpendicular to 
the plane of the couple, the 
length of the vector represent- 
ing the moment of the couple. 
It is agreed that the vector 
shall point from the plane toward that side from which 
the rotation appears counter-clockwise. 




Fig. 79 



78 



APPLIED MECHANICS FOR ENGINEERS 



Since the couple may be moved anywhere in its plane 
or in a parallel plane, the vector may be laid off from any 
point. Thus the vector AB may represent any one of 
the three couples shown in Fig. 79.* 

The theorem of Art. 48 expressed in terms of vectors is : 
ITie projection of the vector representing a couple upon any 
straight line represents the moment of the couple with respect 
to that line. The theorems of Arts. 46 and 49 expressed 
in terms of vectors are as follows : Couples may he com- 
bined hy adding the vectors representing them; and any 
couple may he replaced hy two or more couples whose added 
vectors give the vector of the original couple. 

51. Rectangular Components of a Couple. — A case of par- 
ticular interest coming under the theorem just mentioned 




X 



/ 


v-V. 






1 












u 




mI 


1 


hh 






1 



Fig. 80 

is the combination of couples in three rectangular planes 
into a single couple and the converse problem of resolving 
a couple into component couples in three rectangular 

* The arrow barb is placed a short distance from the end of the couple 
vector to distinguish it from the force vector. 



COUPLES 



79 



planes. Let the planes be taken as coordinate planes. 
Call the moments of the couples in the ys-plane, the xz- 
plane, and the a;2/-plane, M^^ My^ and M^ respectively. 
Representing the couples as vectors, the values M^^ My^ 
and Mz are laid off along the a:-, ?/-, and 2-axes respectively 
(Fig. 80). The vector representing the resultant couple 
is then the diagonal of the rectangular parallelopiped 
with these vectors as edges. 

If the moment of the resultant couple is M and the 
direction angles of the vector representing the couple are 
\, /x, and v^ then 

M= VJ// + MJ' + M?, 
. M, My M, 

cos X = —4, cos IX = — -^, COS V = —i. 

M M M 

Conversely, if X, /x, v^ and M are given, the component 
couples in the yz-. xz-^ and a;^-planes have moments equal 
respectively to M^ = il[f cos X, My = il[f cos /x, M^ — Mcos v. 

52. A Couple not Balanced by a Force. — A force and a 
couple cannot be in equilibrium. Let P and P' form a 
couple and let Pj be a force. No 
matter where the force P^ may act 
the couple may be moved in its own 
plane or in a parallel plane until one 
of its forces, say P, has at least one 
point in common with the force Pj. 
The forces P and P^ then either 
annul each other or they have a 
resultant passing through their common point. 




Fig. 81 



This 



resultant and P' cannot balance each other, for two forces 




80 APPLIED MECHANICS FOR ENGINEERS 

can balance only when they have the same line of action 
and are equal and opposite. Hence the force and the 
couple cannot balance each other. 

53. Substitution of a Force and a Couple for a Force. — Let 

P be any force acting at a point A and any other point 

distant d from the line of 
P^/^ P' At put in two equal 

and opposite forces P and 
P' parallel to the force P 
at A. Then P at ^ and 
P' at form a couple 

Ftp 82 

whose moment is Pd^ and 
there remains the force P at 0. 

Hence any force may be replaced by a force equal and 
parallel to the given force acting at any desired point, and 
a couple in the plane of the given force and the point 
whose moment is equal to the moment of the given force 
with respect to that point. 

Problem 83. Replace a force of 20 lb. acting in a line 1-5 in. from a 
point by a force acting at and a pair of forces in parallel lines 6 in. 
apart in a plane parallel to the plane of the given force and the point 0. 

Problem 84. Parallel and opposite forces of 50 lb. each act in 
lines distant 20 in. apart in a plane the normal to which has di- 
rection angles A = 30°, /z = 50°, v = ? . The normal passes through 
the first octant, and the moment of the couple appears negative to an 
observer in the first octant if the plane of the couple is regarded as 
passing through the origin. Find the moments of this couple with 
respect to the coordinate axes. 

Problem 85. Determine the magnitude and plane of action of 
the resultant couple of the couples acting on the rectangular block 
shown in Fig. 83. 



COUPLES 



81 



10" 







20 LBS, 



12 LBS. 



LBS. 



10 LBS. 



12" 




Problem 86. Forces of 10, 15, and 20 lb. act respectively in the 
XZ-, XT/-, and yz-planes, parallel respectively to the z-, x-, and 2/-axes, at 
distances respectively of 15, 12, 
and 10 in. from the origin. Re- 
place these forces by a single force 
acting at the origin and a couple. 
Find the magnitude and direction 
of the force vector and of the 
couple vector. 

Problem 87. Two forces, each 
equal to 10 lb., act in a vertical 
plane so as to form a positive 
couple. The distance between the j 
forces is 2 ft. Another couple 
whose moment is equal to 80 lb. -in. 

acts in a horizontal plane and is negative. Required the resultant 
couple, its plane, and direction of rotation. 

Problem 88. A couple whose moment is 10 lb. -ft. acts in the xy- 
plane ; another couple whose moment is — 30 lb. -ft. acts in the xz- 
plane and another couple whose moment is — 25 Ib.-ft. acts in the 
?/2;-plane. Required the amount, direction, and location of the couple 
that will hold these couples in equilibrium. 

Problem 89. Show that the moment of a couple with respect to 
an element of a right circular cone, whose axis is perpendicular to the 
plane of the couple, is the same for all elements of the cone. 



Fig, 83 



6 



CHAPTER VI 

NON-CONCURRENT FORCES 

54. Forces in a Plane. — The most general case of forces 
in a plane is that one in which the forces are non-concur- 
rent and non-parallel. We shall now consider such a 

Y 




case. Let the forces- be Pj, P^, Pg, P4, etc., as shown in 
Fig. 84, and let them have the directions shown. Select- 
ing arbitrarily an origin and a pair of rectangular axes in 
the plane of the forces, replace each force by an equal 

82 



NON-CONCURRENT FORCES 83 

and parallel force acting at the origin, and a couple 
whose moment is equal to the moment of the force with 
respect to the origin (Art. 53). The forces acting at the 
origin may then be replaced by a single force, 



i^ = V(2X)2 + (2r)2. 

The couples, being all in one plane, are equivalent to a 
single couple in that plane whose moment is 

M= l>Pd. (Art. 46.) 

Therefore the system of non-concurrent forces in a plane 

may be reduced to a single force, H = V (^Xy + (2 1^)^, 
acting at an arbitrarily selected origin, and a single 
couple in the plane of the forces whose moment is the 
sum of the moments of the forces with respect to that 
origin. 

For equilibrium, lt = and ^Pd = 0, or SX = 0, SI" = 0, 
and SjPci = ; that is, for equilibrium, the sum of the 
components of the forces along each of the two axes is 
zero and the sum of the moments with respect to any 
point in the plane is zero. Conversely, if IX = 0, ^Y=0, 
and ^Pd = 0, for a point in the plane, the resultant force 
M and the resultant couple both vanish and the forces are 
in equilibrium. 

55. Other Conditions for Equilibrium of Forces in One 

Plane. — The forces may be combined in succession until 
there is obtained either a final resultant, including a zero 
resultant, or a couple (Arts. 13 and 25). In either case 
the sum of the moments of the forces with respect to 



84 APPLIED MECHANICS FOR ENGINEERS 

any point in the plane is equal to the moment of their 
resultant force or couple (Arts. 22 and 46). If the 
sum of the moments of the forces with respect to any 
selected point is zero, the forces cannot be equivalent 
to a couple, for the moment of a couple is the same for 
all points of the plane. The forces may, however, have 
a resultant force. But if in addition the sum of the 
moments of the forces about two other points of the plane 
not in the same straight line with the first point is zero, 
the resultant must be zero ; for either the resultant or 
the arm of the resultant must be zero, and the arm can- 
not be zero for three different points not in the same 
straight line. 

Hence, if the sum of the moments of a set of forces in 
one plane with respect to three points not in the same 
straight line is zero, the forces are in equilibrium. 

It is left as an exercise for the student to prove that 
if the sum of the rectangular components in one direction of 
a set of forces in one plane is zero and the sum of the 
moments of the forces with respect to each of two points in 
the plane not in a perpendicular to the given direction is 
zero., the forces are in equilibrium. 

Problem 90. Prove the statement of the last paragi'aph. 

Problem 91. The following forces in one plane act upon a rigid 
body: a force of 100 lb. whose line of action makes an angle of 45° 
with the horizontal, and whose distance from an arbitrarily selected 
origin is 2 ft. ; also a force of 50 lb. whose line of action makes 
an angle of 120"^ with the horizontal, and whose distance from the 
origin is 3 ft. ; and a force of 500 lb. whose line of action makes an 
angle of 300° with the horizontal and whose distance from the origin 
is 6 ft. Find the resultant force and the resultant couple. 



NON-CONCURRENT FORCES 



85 



1 TON 




Fig. 85 



X 



Problem 92. It is required to find the stress in the members 
AB, BC, CD, and CE of the bridge truss shown in Fig. 85. 

Note. The member 
AB is the member be- 
tween A and B, the 
member CD is the mem- 
ber between C and D, 
etc. This is a type of 
Warren bridge truss. 
All pieces (members) 
are pin-connected so 
that only two forces act 

on each member. The members are, therefore, under simple tension 
or compression; that is, in each member the forces act along the piece. 
Solution of Problem. The reactions of the 
supports are found by considering all the external 
forces acting on the truss. Taking moments 
about the left support, we get the reaction at the 
right support, equal to 4500 lb. Summing the 
vertical forces or taking moments about the right- 
hand support, the reaction at the left-hand support 
is found to be 3500 lb. 
Cutting the truss along ah and putting in the forces exerted by the 
left-hand portion, consider the right-hand portion (see Fig. 86). The 
forces C and T act along the pieces, 
forming a system of concurring 
forces. For equilibrium, '^X = d 
and 2F = 0, giving two equations, 
sufficient to determine the un- 
knowns C and T. The forces in 
the members CD and CE may now 
be considered known. 

Cutting the truss along the line 
cd and putting in the forces exerted by the remaining portion of 
the truss, we have the portion represented in Fig. 87. This gives a 




\l 



4500 LB 

Fig. 86 



t 



2 TONS 




Fig. 87 



86 



APPLIED MECHANICS FOR ENGINEERS 



system of concurring forces of which C and 2 tons are known, so 
that from the equations '^X =0 and 2F = the remaining forces C^ 
and T^ may be found. 

Problem 93. In the crane shown in Fig. 88 (a) find the forces 
acting on the pins and the tension in the tie AC. The method of 
cutting cannot be used in this case since the vertical and horizontal 
members are in flexure. Taking the horizontal member and consid- 
ering all of the forces acting upon it, we have the .system of non-con- 
curring forces shown in Fig. 
88(6). Three unknowns are 
involved, Pg, 1\, and Pg, and 
these may be determined by 
three equations "^X = 0, ^y 
= 0, and 2P</ = 0. It is to 
be remembered that the pin 
pressure at E is unknown in 
magnitude and direction. In 
all such cases it is usually 
more convenient to resolve 
this unknown pressure into 
its vertical and horizontal 
components, giving two 
unknown forces in known 
directions instead of one un- 
known force in an unknown 
direction. This will be done 
in all problems given here. 
In the present case the two 
forces P^ and Pg are the components of the unknown pin pressure. 

The tension in the tie AC may be found by considering the forces 
acting on the whole crane and taking moments about B. Thus 
2P/^/ = gives, calling the tension in the tie T, 

r35 sin 45^ = 8000 (25), 
y ^ 8000 (25) 
35 sin 45°' 




I 



(b) 



20- 



-5^ = 



Fig. 88 A 



or 



NON-CONCURRENT FORCES 



87 



Problem 94. In 

the crane shown 
in Fig. 89 (a) find 
the forces in the ties 
and the compres- 
sion in the boom. 
The method of cut- 
ting may be used 
here to determine 
the compression T 
and the compres- 
sion in the boom, 
since ^^ is not in 
flexure, if we neg- 
lect its own weight. 
Cutting the struc- 
ture about the point 
A and drawing the 
forces acting on 
the body, we have 
the system shown in 




Fig. 89 (b). 




Fig. 90 



Fig. 89 

The forces W may be considered as 
acting at the center of the pulley. 
The system of forces is concurring, 
so that 2X = 0, and ^Y = are 
sufficient to determine T and C. 
T' may be found by considering 
the forces acting on the whole 
crane and taking moments about 
the lowest point. What length of 
AB would make the stress, T, ten- 
sion? 

Note. Neglecting friction, the 
tension W in the cord supporting 
the weight is transmitted undimin- 
ished throughout its length. 



88 



APPLIED MECHANICS FOR ENGINEERS 



Problem 95, Find the horizontal and vertical components of the 
forces acting on the pins of the structure shown in Fig. 90. 

Suggestion. First take the vertical member and consider all the 

forces acting on it. 

Problem 96. Find the 
forces acting on the pins of 
the structure shown in Fig. 
91, the weight of the mem- 
bers AD, BF, and CE being 
600 lb., 400 lb., and 100 lb., 
respectively. 

Problem 97. A traction 



engine is passing over a 
bridge, and when it is in the 
position shown in Fig. 92, one half of the load is carried by each truss. 
The weight of the engine is transmitted by the floor beams to the 
cross beams, and these are carried at the pin connections of the truss. 
Find the stress in the members AB, BC, CE, CD, and DF, for the 
position of the engine shown. The weight of the engine is 3 tons. 

Note. The floor beams are supposed to extend only from one 
cross beam to another. , 





Fig. 92 



NON-CONCURRENT FORCES 



89 



Problem 98. In Problem 94, suppose the weight of the boom to 
be one ton ; find the stresses T and T' and the pin pressures. 

Note. The boom is now under flexure, so that the method of cut- 
ting cannot be used. 

Problem 99. A dredge or steam shovel, shown in outline in Fig. 
93, has a dipper with capacity of 10 tons. When the boom and dipper 
are in the position shown, find the forces acting on AB, CD, and EF. 
The projection of the point i*' is 6 ft. from the point E. 




Mo TONS 



Fig. 93 



Suggestion. Consider fi.rst all the forces acting on CD, then all 
the forces acting on AB. 

Note. The member EF has been introduced as such for the sake 
of analysis ; it replaces two legs, forming an A frame. 

Problem 100. Suppose the members of the structure in Problem 
99 to have weights as follows : AB, 15 tons, and CD, 3 tons, not in- 
cluding the 10 tons of dipper and load. Find the forces as required 
in the preceding problem. 

Problem 101. Suppose the beam in Problem 5 to be 20 ft. long 
and to have a weight of 2000 lb. ; find the pin reaction and the ten- 
sion in the tie. 

Problem 102. Assume that the compression members of the 
Warren bridge truss of Problem 92 have each a weight of 500 lb. ; 
find the stresses in the members BC and CE. 



90 



APPLIED MECHANICS FOR ENGINEERS 



56. Forces in Space. — First consider a single force P 
acting at A (Fig. 94). Choose any point and rectan- 
gular axes through as origin. Let the coordinates of 




Fig. 94 



A referred to these axes be (a:, ^, 2) and let the components 
of P parallel to these axes be X, Z, Z. Construct a rec- 
tangular parallelopiped with and A as opposite vertices 
and with x^ y^ and z as edges. Along these edges put in 
two pairs each of parallel, equal, and opposite forces, 
parallel to JT, Z", and Z respectively, one force of each set 
acting at in the direction of the corresponding compo- 
nent at A (Fig. 94). It is then evident, from an inspection 
of the figure, that the force P acting at ^ is equivalent 
to the components X, y, Z acting at and three couples 
as follows : 



NON-CONCURRENT FORCES 



91 




Fig. 95 



in the ^2-plane a couple whose moment is zY — yZ^ 
in the a^2-plane a couple whose moment is xZ — zX^ 
in the :z;?/-plane a couple whose moment is yX— xY. 

In determining the sign of the moment the observer is 
supposed to look toward the 
origin from the positive end 
of the axes (Fig. 95). 

From the above discussion of 
a single force it follows at once 
that any number of forces in 
space are equivalent to three 
forces, 

2X, SF, 2Z, 

acting along rectangular axes 

at an arbitrary point, and three couples in the ^/2-, xz-^ and 
ici/-planes whose moments are respectively 
M, = ^(zY- yZ), My = ^(xZ-zX), M, = ^^iyX-xY). 
The three forces acting at the origin may be combined 
into a single resultant, 

B = ^(IXy + (2 F)2 + (SZ)2, 
whose direction cosines arc 

cos a = ——, cos p = -— -, cos 7 = ——' (Art. -U.J 
R Ji Jt 

The couples may be combined into a single couple of 
moment 

whose direction cosines are 



cos X = 



M' 



cos \x 



M^ 
M' 



cos V = 



M 



(Art. 51.) 



92 



APPLIED MECHANICS FOR ENGINEERS 



For equilibrium both M = and M= ; that is, 

2X=0, 2:F=0, 2Z=0, il^, = 0, i!f, = 0, i!f, = 0. 

Since zY—yZ is the moment of the force P with 
respect to the a;-axis, and hence M^ is the sum of the 
moments of all the forces with respect to the r?;-axis, the 
above conditions may be expressed in the words: the sum 
of the components of the forces along each of the three arbi- 
trarily chosen axes is zero^ and 
the sum of the moments with re- 
spect to each of these axes is zero 
when the forces are in equilibrium. 

Problem 103. A vertical shaft is 
acted upon by the belt pressures T^ 
and 7*2, the crank pin pressure, P, 
and the reactions of the supports. (See 
Fig. 96.) Write down the six equa- 
tions for equilibrium. 

Note. The ?/-axis has been chosen 
parallel to the force P, and T^ and 7*2 
are parallel to the a:-axis. 

2.Y = .Y' + X" - T^- T, = 0, 
2F=: F + Y"- P = 0, 

^Z = Z" - G~G' = 0] 
M^ = - Pb - Y"l = 0, 



M, = X"l - T,c - T,c- G'l = 0, 

M, =Pa-\- T^r - T^r = 0. 

From these six equations six unknown 
quantities can be found. If G, G', 
Ti, and T2 are known, the reaction of the supports and P may be fpund. 

Problem 104. In Problem 103 suppose a = V, h = 8", c = 6", 
I = 3', r = Q", G = 90 lb., G' = 40 lb., T, = 75 lb., T^ = 200 lb.; write 




NON-CONCURRENT FORCES 



93 



the equations for equilibrium and solve for P and the reactions of the 
supports. 

Problem 105. Same as Problem 104 except that T-^ and T^ make 
an angle of 30"^ (counter-clockwise when observed from top) with 
their directions, as shown in Fig. 96. 

Problem 106. A crane shown in Fig. 97 has a boom 45 ft. long 
and a mast 30 ft. high. It is loaded with 20 tons, and the angle 
between the boom and the mast is 45°. The two stiff legs each make 
all angle of 30° with the mast and an angle of 90"^ with each other. 
Find the pin 
pressures in 
boom and mast, 
also the stress in 
the legs when 
(«) the plane of 
the crane bisects 
the angle be- 
tween the legs 
and (6) the plane 
of the crane 

makes an angle of 30° with one of them. If the boom weighs 4000 lb., 
find the stress in the legs when the plane of the crane bisects the angle 
between them. Assume that the pulleys A and B are at the ends of 
the boom and mast respectively. 

Problem 107. Suppose the shaft of Problem 103 to be horizontal, 
find P and the reactions of the supports. Assume y horizontal and 
perpendicular to the shaft, and x vertical. 

57. Graphical Method for Forces in One Plane. — Let P^, 

P^, Pg, P^ (Fig. 98) be a set of forces in one plane ; to 
find by graphical processes a force P^ that will balance 
them. 

Resolve P^ into two components, S^^ g ^^^L Sr^^ ^ in arbi- 
trary directions. Extend the line of S^^^ until it meets 




Fig. 97 



94 



APPLIED MECHANICS FOR ENGINEERS 



the line of Pg* Resolve P^ into two components, S-^^^^^ 
equal and opposite to aS'j 2 <^>^ J^v '^"^^ ^^2,3- Extend the 
line of aS2,3 to cut the line of Pg and proceed as before. 

When the last force, P^, 
is reached, there remain 
unbalanced only the 
component, iS'g ^ of the 
first force Pj, and )S^^ 5 of 
the last force P^. The 
force Pg that will bal- 
ance these components 
will balance the original 
forces. Hence Pr is de- 
termined in magnitude, 
direction, and line of 
action as the force that will balance these two compo- 
nents (Fig. 98). 

The force Pg is called the equilihrant of the forces 
P, 




Fig. 98 



A- 



^5, with vertices on the lines of 



The polygon A-^A^ 
the forces, is known as an equilibrium polygon. If the 
sides of this polygon were replaced by Aveightless rods, or 
links, hinged at the vertices, the forces P^ ••• P^ would be 
just balanced by the tensions (or compressions) in the 
rods, and the framework would retain its position. An 
indefinite number of equilibrium polygons may be con- 
structed for a set of balanced forces. 

A shorter method of constructing the equilibrium poly- 
gon and determining the force, Pg, to balance the given 
forces is the following: One triangle from each vertex 



NON-CONCURRENT FORCES 



95 




Fig. 99 



^j, '" Ar^ may be taken and their common sides put 
together, without changing their directions, to form a 
closed polygon whose sides are the vectors of the forces 
laid off in succession (Fig. 99). This polygon is called 
the force polygon. From the force polygon it is evident 
that the vector of the force Pg, the 
e(][uilibrant of the given forces P^ ••• 
P^, is the closing side of the polygon 
formed by laying off in succession the 
vectors of the given forces, the begin- 
ning of each vector being placed at the 
end of the preceding one. Having 
laid off the force polygon and thus de- 
termined the magnitude and direction 
of the equilibrant, its line of action is determined by 
drawing the equilibrium polygon ; but this may now be 
done without constructing the triangles at the vertices 
-4 J, A^ •". For the lines (called rays) from the point 
in the force polygon are parallel to the corresponding 
sides of the equilibrium polygon, and since the directions 
of /S'j 2 ^11^^ ^5,1 were arbitrary, the point may be any 
point in the plane. Join to the vertices of the force 
polygon and then construct the equilibrium polygon by 
drawing its sides parallel to the corresponding rays of the 
force polygon ; i.e. a side of the equilibrium polygon 
terminating on two lines of force must be parallel to the 
ray of the force polygon drawn to the intersection of 
those same two forces. 

To sum up: To determine graphically the equilibrant 
of a set of forces in one plane, construct a polygon of the 



96 



APPLIED MECHANICS FOR ENGINEERS 



vectors of the given forces. The closing side is the 
vector of the equilibrant. Draw rays from any point in 
the plane to the vertices of the force polygon. Construct 
the equilibrium polygon by drawing lines parallel to the 
corresponding rays of the force polygon, intersecting on 
the lines of action of the given forces. The intersection 
of the two sides of the equilibrium polygon parallel to 
the rays to the closing side of the force polygon is a 
point on the line of action of the equilibrant. The equi- 
librant is thus completely determined. 

In lettering the rays of the force polygon and the sides 
of the equilibrium polygon it is sufficient to mark them 
only with the subscripts of the forces that they connect. 
Figure 100 illustrates this. 





Fig. 100 



When the forces are in equilibrium, as P^, Pg^ ••* ^5 i^ 
Fig. 100, both the force polygon and the equilibrium 
polygon are closed. It is possible to have the force poly- 
gon close without having the equilibrium polygon close ; as. 



NON-CONCUBBENT FOBCES 



97 



for example, P^^ ••• ^4 and a force P^ parallel and equal 
to the closing side of the force polygon but not passing 





Fig. 101 

through the remaining vertex of the equilibrium polygon 
(Fig. 101). The forces in this case form a couple, for the 
forces Pp •••P4 are equivalent to a force equal and oppo- 
site to Pg passing through the vertex A^ of the equilibrium 
polygon. 

Problem 108. Determine graphically the equilibrant of the four 
forces of Fig. 102. Where 
does it cut the line J.5? 



Problem 109. Find 
graphically the reactions 
due to the loads in Fig. 103. 

Suggestion. The force 
polygon here becomes a 
straight line. Call the re- 



1000 LBS. 



500 LBS, 



jAl 



A 



h^'- 



2000 LBS. 



1500 LBS. 



^' 




IJ 



Fig. 102 



actions P^ and P., and agree 

that P4 shall follow Ps in 

the force polygon. P. must then close the polygon. The missing ray 

to the intersection of P^ and P^ is found by drawing from a ray par- 



98 



APPLIED MECHANICS FOB ENGINEERS 



allel to the closing side (dotted) of the equilibrium polygon. Hence 
7^4 and P- are determined. Check by moments. 





Fig. 103 



When a truss is acted upon by wind loads, the directions 
of the supporting forces depend upon the manner in which 
the truss is attached to the supports. Some trusses are 
pinned at one end and rest on rollers at the other. The 
reaction at the roller end may then be assumed to be 
vertical, the pins taking all of the horizontal load. If the 
truss is fixed at both ends, the reactions may be assumed 

to be in parallel lines, or 
else that the supports 
take equal parts of the 
horizontal load. 



500 LB. 




Fig. lOi 



AB in Fig. 104 carries loads as shown. 



Problem 110. The beam 
Find graphically the reactions 



at A and B, given that the reaction at B is vertical. 



NON-CONCURBENT FORCES 99 

Suggestion. Since the point A is the only known point on the 
reaction through ^, begin the equilibrium polygon at A as a vertex. 

Problem 111. Find graphically the reactions of the truss of 
Fig. 105, assuming that 

the reactions are in I ^ ''■°'^^' 

parallel lines. 

Problem 112. Find 
graphically the reac- 
tions of the truss of 
Fig. 105, assuming that 
the left-hand reaction 
is vertical. 



2 TONS 




Fig. 105 



58. Stresses in Frames. — Stresses in roof and bridge 
trusses are usually computed on the supposition that the 
members are two force pieces; i.e. are pin-connected at 
the ends and have loads applied only at the joints. The 
stresses in the truss of Problem 92 were computed on that 
supposition. The graphical method of determining the 
stresses is often more easily and quickly applied than the 
analytical method, except for very simple cases. As an 
illustration of the graphical method, the stresses in the 
truss of Fig. 106 are determined. 

It is convenient here to use Bow's notation. Represent 
a force acting on the truss, or a member of the truss, by a 
pair of letters, one on each side of the line of the force or 
member. Thus the left-hand load on the truss is read 
AB^ the right-hand reaction is CD, the upper horizontal 
member of the truss is BF, etc. Determine the reactions, 
either graphically or by moments. They are CD = -^^s 
DA = ^-. The loads and reactions form a system in 
equilibrium. These forces are laid off on a vertical line, 



100 



APPLIED MECHANICS FOR ENGINEERS 



known as the load line, in succession in the order in which 
they are met in going around the outside of the truss, as 
AB, BO, CD, DA, The force AB is represented on the 
load line by the same letters in small type, ab, etc. At 
any joint of the truss three or more forces act in the 



5 TONS. 





Fig. 106 



directions of the forces and members that meet at that 
joint. These forces are in equilibrium and must therefore 
form a closed polygon when laid off in succession (Art. 
16). In constructing the polygons the forces already laid 
off on the load line are made use of. Also a polygon of 
forces, all of which are known in direction, cannot be 
constructed if more than two forces are unknown in 
magnitude. Hence the force polygon is first drawn for a 
joint where only two forces are unknown. The determi- 
nation of the forces acting at this point gives additional 
known forces at other points, for the force in any member 
acts equally and in opposite directions on the pins at its 
ends. 



NON-CONCURRENT FORCES 101 

Choosing the joint of the truss at the left support, the 
forces DA, AE, and ED must form a closed triangle. 
Hence using the force da on the load line, complete the 
force triangle dae by drawing through d and a lines 
parallel respectively to DE and AE to meet in e. The 
lines ae and ed represent the forces in AE and EB. The 
directions of the forces acting at the joint are those 
given by passing around the triangle dae in the order 
d-a-e, since the force da acts from d towards a. Next go 
to the joint ABFE and construct the force polygon for 
that joint, using the known forces ea and ah, drawing 
lines through e and h parallel respectively to EF and BF 
to intersect in /. The polygon ahfea is then the force 
polygon for the joint ABFE. The directions of the 
forces acting at the joint are given by passing around the 
polygon in the order a-h-f-e-a, since the force ah acts from 
a toward h. The direction in which the forces act at the 
joint are indicated by putting arrows on the members 
near the joint. Thus the member AE pushes against the 
pins at its ends, while EB pulls on the pins at its ends. 
AE is in compression, EB in tension. 

The whole stress diagram (Fig. 106 (5)) is thus con- 
structed upon the load line. The stress in any member is 
represented by the line in the force diagram terminating 
in the same letters as those of the member, as^ represents 
the force in FGr. 

A check on the correctness of the work is that the last 
line of the force diagram must be parallel to the cor- 
responding member of the truss. More accurate results 
are obtained by drawing the figure to large scales. 



102 



APPLIED MECHANICS FOR ENGINEERS 



Problem 113. In the truss of Fig. 107 the apex is distant ^ the 
length of the span above the lower chord. The panels carry equal 
loads, half the load of each panel being transferred to the truss at the 




Fig. 107 

pin at the end of that panel. Calling the total load unity, find graphi- 
cally the stresses in the members of the truss. Write the values of 
the stresses on the figure of the truss and indicate whether tension 
or compression. 

Problem 114. A wind load acts on the truss of Fig. 107, as shown 
in Fig. 108. Assuming the reactions to be in parallel lines, find the 
reactions and stresses in the members graphically, and write down 
the stresses as in the preceding problem. 

'A 




Fig. 108 



Suggestion. In finding the reactions the work will be simplified 
by combining the loads into one load. 



NON-CONCUEEENT FOECES 



103 




Problem 115. Assuming 
the total wind load to be equal 
to I of the total vertical load 
in the preceding problems find 
the stresses when both loads 
act simultaneously. 

Problem 116. Find the 
stresses in the truss of Fig. 
109, all members of the truss being of the same length. 

2 TONS. 2 TONS. Problem 117. Find the 

stresses in the members of 
the cantilever truss of Fig. 
110. 

Problem 118. Find the 
stresses in the truss of Fig. 
111. 

59. Method of Substi- 
tution. — It sometimes 
happens that in con- 



FiG. 110 

structiog the force diagram a point is reached where it 



1 TON 




1 TON. 



3 TONS. 



3 TONS 



3 TONS. 




Fig. Ill 



104 



APPLIED MECHANICS FOR ENGINEERS 



is not possible to proceed to another joint at which 
there are only two unknown forces acting. The temporary 
removal of two members and the substitution of another 

member to hold the 
truss from collapsing 
will sometimes en- 
able one to continue 
the construction of 
the force diagram. 
This is illustrated in 
the case of the Fink 
truss shown in Fig. 
112. 

Starting at the left 
support, the force 
polygons fag, alhg, 
and fghk can be con- 
structed. The poly- 
gons for the remain- 
ing joints in the left 
half of the truss then 
have three sides miss- 
ing and cannot be completed in the usual way. If the 
members ML and MN be removed, and a member M N 
(Fig. 112(5)) be substituted, the truss would stand. 
Moreover, the stresses in Z>iV, NE^ and EF would not 
be changed ; for if sections were made across these 
members in the two cases, the parts of the truss to the 
left of the cuts would be acted upon by the same forces, 
and hence would require the same forces in BN, NE^ and 




NON-CONCUBBENT FOBCES 105 

UF to balance them. (The same is true for sections 
across BIT, HK^ and KF.) With the changed form of 
the truss the force polygons hhhcmJ and m' cdn can be con- 
structed. Having located w, we may return to the origi- 
nal truss and construct the polygon nmcd^ then hcmlkh^ etc. 

Problem 119. The panels of a Fink truss are equal and carry 
equal loads. The upper chord is inclined 30° to the horizontal, and 
the lower chord is horizontal. Calling the total load unity, construct 
the stress diagram and write the stresses in the members on the figure 
of the truss, indicating tension and compression. 



CHAPTER VII 

MOMENT OF INERTIA 

60. Definition of Moment of Inertia. — The study of 
many problems in mechanics brings to our attention the 

value of the integral of the form | y'^dF^ where dF repre- 
sents an infinitesimal area and y is the distance of that 
element from an axis of reference. The value of this 
integral taken over a given area is called the moment of 
inertia of that area with respect to the given axis of ref- 
erence. In like manner j r^dV and j r^dM taken to in- 
clude all elements of a given volume or a given mass are 
called respectively the moment of inertia of the volume, 
and mass, with respect to the given axis. In these inte- 
grals dV and dM represent infinitesimals of volume and 
mass respectively, and r the distance of the infinitesimal 
element from the axis of reference. We shall designate 
moment of inertia by the letter /. Thus we write : 

I^jyHF, 

I=fr^dV, 

for area, mass, and volume, respectively. 

106 



MOMENT OF INERTIA 107 

Many problems that confront the engineer involve in 
their solution the consideration of the moment of inertia. 
This is the case when the energy of a rotating flywheel, 
for example, is being determined. The energy of a rotat- 
ing body (Art. 137) is expressed as follows : 

Kinetic energy = -^, 

LA 

where / is the moment of inertia with respect to the axis 
of rotation and to is the angular velocity. (See Art. 118.) 
It is seen that the energy of rotating bodies, having the 
same angular velocity, or the same speed, is directly pro- 
portional to their moments of inertia. The quantity, 
therefore, plays a very important part in the considera- 
tion of rotating bodies. 

Also the strength of a beam or column depends upon 
the moment of inertia of the area of the cross section of 
the beam or column with respect to a line of the section 
through the center of gravity of the section. 

In a later chapter a reason will be seen for the name 
" moment of inertia." For the present it may be regarded 
as a name arbitrarily applied to a quantity frequently 
used in the applications of mechanics. (See Art. 137.) 

61. Radius of Gyration. — The moment of inertia of an 
area involves area times the square of a distance. We 

may write 1= \ y'^dF = Fk"^^ where F is the area and k is 

a distance, at which, if the area were all situated, the 
moment of inertia would be unchanged. This distance 
k is called the radius of gyration. In a similar way 



108 APPLIED MECHANICS FOR ENGINEERS 

for a mass we write : 1= j yMM== Mk^, and for volume 
1= Cy2dV=^Vk\ 

62. Units of Moment of Inertia. — The moment of inertia 
of an area with respect to any axis may be expressed 
as Fk'^. The area involves square inches, and k"^ is a dis- 
tance squared. The product is expressed as inches to the 
fourth power. The moment of inertia of a volume Vk^ 
requires inches to the fifth power. The moment of in- 
ertia of a mass requires in addition to Vk"^ the factor ^, 

g 

so that pounds and feet per second per second are involved. 
This is somewhat more complicated since it involves units 
of weight, distance, and time. The presence of ^ (= 32.2) 
in the expression requires that all distances be in feet. 
It is customary to express the moment of inertia of a mass 
without designating the units used, it being understood 
that feet, pounds, and seconds were used. 

63. Representation of Moment of Inertia. — From the defi- 
nition of moment of inertia it is evident that an area has 
a different moment of inertia for every line in its plane. 
We shall designate the moment of inertia with respect to 
a line through the center of gravity by Ig with a subscript 
to indicate the particular gravity line intended. For 
example, Ig^ indicates the moment of inertia with respect 
to a gravity axis parallel to x, and Igy indicates tlie 
moment of inertia with respect to a gravity axis parallel 
to y. The moment of inertia with respect to a line other 
than a gravity line will be designated by /, the proper 



MOMENT OF INERTIA 



109 



subscript indicating the particular line. Similar sub- 
scripts will be used to designate the corresponding radii 
of gyration. It should be noted that moment of inertia 
is not a quantity involving direction. It has to do only 
with magnitude and is essentially positive. 

64. Moment of Inertia. Parallel Axes. 




(a) For Areas. In Fig. 113 (a), let OX and 0' X' be 
two parallel axes in the plane of the area F. If y is 
the ordinate of an element of area dF referred to 0' X 
and h the distance between the axes, then 

I^ = ^(y + },yclF=j^{y'^ ■Y'lhy^ W)dF 

= CyHF + 2 Ji^ydF + h^^dF 
= r,-{-2.hyF-^h'^F, 

where y is the ordinate of the center of gravity of the 
area Preferred to the axis 0' X' (Art. 36). 

If 0' X passes through the center of gravity of #, then 
y is zero and the equation may be written 



110 APPLIED MECHANICS FOR ENGINEERS 

where Ig is the moment of inertia of F with respect to a 
gravity axis parallel to OX. This relation may be 
expressed in words as, 

The moment of inertia of an area with respect to any line 
in its plane is equal to its moment of inertia with respect to 
a parallel gravity axis plus the area times the square of the 
distance between the two axes. 

(b) For Solids. In Fig. 113 (5), OX and O'X' are 
parallel axes distant h apart. dM is the mass of an ele- 
ment of the body distant r from OX and r' from O'X' . 
If the coordinates of dMsive (^, s), as shown in the figure, 
then 

= h^-\-2hy-h r''\ 

Therefore I, = Cr^dM=f(h^ + 2 % + r'^)dM 

= h^CdM-{- 2 hCydM+fr'^dM 

= h^M+2hyM+r,. 

If 0^ X' passes through the center of gravity of the solid, 
y = and the equation reduces to 

From the above formulae it is evident that the moment 
of inertia of an area or a solid about a gravity axis is less 
than that for any parallel axis. 

65. Moment of Inertia ; Inclined Axis. — It is often desir- 
able, when I^ and ly are known, to find the moment of inertia 
with respect to an axis w making an angle a with x. (See 



MOMENT OF INERTIA 



111 




Fig. 114.) Here, 7, = Cv^dF and I, = Cw^dF. lu terms 
of x^ y, and a, 

Iy,= j (7/ cos a — X sin a)'^dF 

= I y2 cos^ a c?7' —^\xy cos a sin a c?^^ + \ ^ sin^ a c?JP 
= cos^ a I ?/2<^jP — 2 sin a cos a I xydF + sin^ « | aj^c?^ 
= I^ cos^ a — sin 2 « j xydF + i^ sin^ a. 

In a similar way 

ly = /j. sin^ a + 2 sin a cos a I xydF + /^ cos^ a, 

where OF is perpendicular to OW. 

These are the required formulse for obtaining the mo- 
ment of inertia with respect to inclined axes. It follows 
that 

^w ~(" -'-V — J-x + -*»/• 



112 APPLIED MECHANICS FOR ENGINEERS 

That is, the sum of the moments of inertia of an area 
with respect to two rectangular axes in its plane is the 
same as the sum of the moments of inertia with respect to 
any other two rectangular axes in the same plane and 
passing through the same point. This states that the 
sum of the moments of inertia for any two rectangular 
axes through a point is constant. It will be seen in 
Art. 68 that this constant is the polar moment of inertia. 

66. Product of Inertia, — The integral i xydF is called 

a product of inertia^ for want of a better name. In case 
the area has an axis of symmetry, either the x- or ?/-axis 
may be taken along such an axis. The product of inertia 
then becomes zero, since if x is the axis of symmetry, for 
every + y there is a corresponding — y. A similar rea- 
soning shows that the product of inertia is zero when y is 
the axis of symmetry. In such cases 

and I^ = I^ sin- ^ + ly cos^ a. 

When I xydF is not equal to zero, it is necessary to 

select the proper limits of integration and sum the in- 
tegral over the area in question. This is illustrated in 
Art. 76. 

67. Axes of Greatest and Least Moment of Inertia. — It is 

often important to know for what axis through the center 
of gravity the moment of inertia is least or greatest ; that 
is, what value of a makes i^ or I^ a maximum or a mini- 
mum. For any area J^, ly, and j xydF are constant after 



MOMENT OF INERTIA 113 

the X- and ?/-axes have been selected. Using the method 
of the calculus for finding maxima and minima, we have, 
putting 



/ 



flT 

xydF= ^, ^^^ = (7^ - 7^) sin 2 a - 2 z cos 2 a. 
da 



Equating the right-hand side to zero, the value of a that 
gives either a maximum or minimum is seen to be given 
by the equation 

tan 2 a = — — , 

J-y -/.J. 

or, what is the same thing, 

sin2a= and cos 2 a= ^ 



It is seen upon substituting these values of sin 2 a and 
cos 2 a in 



do? 



= 2(Iy — 7^)cos 2 a + 4 z sin 2 a 



that the positive sign before the radical indicates a mini- 
mum and the negative sign a maximum value for i^. 
The equation for tan 2 a is satisfied by two values of 2 a 
differing by 180°. These values of 2 a correspond to the 
two signs in the values for sin 2 a and cos 2 a. The values 
of a corresponding to the two signs therefore differ by 90°. 
Hence, through any point of the plane, there are two axes 
at right angles to each other about one of which the moment 
of inertia is a maximum and about the other a minimum. 
These axes are known as the principal axes of the area 
through that point. The axes through the center of 



114 



APPLIED MECHANICS FOR ENGINEERS 



gravity of the area for which the moments of inertia was 
greatest and least are known as the Principal Axes of the 
Area. This subject will be further discussed in Art. 76. 

It is seen from the above that when j xydF= 0, the 

values of a which give maximum or minimum values for 
I^ and ly are 0° and 90°. This means that the x- and ?/-axes, 
themselves, are the principal axes. Conversely, if the x- 

and ?/-axes are principal axes, then j xydF = 0. If either 

the X- or ^-axis is an axis of symmetry, 1 xydF = and 
hence the x- and ?/-axes are principal axes. 

Problem 120. Derive the 
formula for 7„ written in Art. 
65. 

Problem 121. Prove that 
when /^ is a maximum, /„ is a 
minimum ; and that when 7^ is 
a minimum, /^ is a maximum, 
by using 

7^ 4- /y = constant. 

68. Polar Moment of In- 
ertia. — The moment of in- 
ertia of an area with re- 
spect to aline perpendicular 
to its plane is called the polar moment of inertia of the area. 
Consider the area represented in Fig. 115 and let the 
axis be perpendicular to the area at any point 0. Let 
dF represent an infinitesimal area and let r be its distance 
from the axis. Representing the polar moment of inertia 




Fig. 115 



MOMENT OF INERTIA 



115 



by ip, we have 






I^^^rUF; 


but 


r^^x^ + ^2^ 


so that 


I^ = jxHF+^y^dF, 



or 



p ~ ^y "^ ^oc' 



That is, the polar momerit of inertia of an area is equal 
to the sum of the moments of inertia of any two rectangular 
axes through the same point. 

It has already been shown that 1^+ Iy = constant (see 
Art. 65) for any point of an area. 

69. Moment of Inertia of Rectangle. — Let it be required 
to find the moment of inertia of the rectangle shown in 
Fig. 116 (a), with respect to the axis x. We may write 



I. 



= fg^dF. 



Since dF = bdg, this becomes 







, 1 * 










4, 




■rn^/z 


d'l 


, 




1 

1 


1 



(a) 



-X 




Fig. 116 



116 APPLIED MECHANICS FOR ENGINEERS 

To find the moment of inertia witli respect to a grav- 
ity axis parallel to x we may make use of the formula 
Ig^ = I^ — Fd'^^ from which we have 

Igx = ^bh^ and fc2^^ = — 

The same result may of course be obtained by taking the 
axes through the center and integrating between the limits 

7 7 

and - • From comparison we may write the moment 

of inertia with respect to a gravity line perpendicular to x^ 

and the polar moment of inertia for the center of gravity 

70. Moment of Inertia of a Triangle. — It is required to 
find the moment of inertia of the triangle shown in Fig. 
116 (^) with respect to the axis x^ coinciding with the 
base of the triangle. We have 

/^ = I if'dF^ where dF = xdy^ 



But a; = - (/i — z/), from similar triangles, giving 
h 



h 
h 



MOMENT OF INERTIA 



117 



The moment of inertia with respect to the horizontal 
gravity axis may now be determined. 

Ig^ = I^- Fd? = gg-, and k% = ^^ • 

The same results may of course be obtained by direct 
integration. 

Problem 122. Find the moment of inertia of tlie area of a 
triangle with respect to an axis through the vertex parallel to the 
base. 

Problem 123. Find the polar moment of inertia of the area of a 
right triangle for the center of gravity. 

71. Moment of Inertia of a Circular Area. — The moment 
of inertia of a circular area Avith respect to a horizontal 
gravity axis x^ as shown in Fig. 117, may be found as fol- 




Fia. 117 



lows : /y^ = I if-dF. Here it is simpler to use polar coordi- 
nates. Changing to polar coordinates, remembering that 



118 APPLIED MECHANICS FOR ENGINEERS 

y = p sin 6, and dF = dp{pdO)^ the integral becomes 

/_= r^ Cp'^sin'^epdedp 

Jo Jq 



^gx 



r4 r*2^ 

= -j 1(1 -COS 2 e)dO 

r'rO 1 . ^.T' 






The corresponding radius of gyration is k^^ = 



ox 2 



On 



account of the symmetry of the figure this is the moment 
of inertia for any line in the plane through the center of 
gravity. It follows that 






and that 



I. 


= 


7rr* 

9 


K 


= 


9 * 



72. Moment of Inertia of Elliptical Area. — Let it be 

required to find Ig^. and Igy of the elliptical area shown 
in Fig. 118. The equation of the bounding curve is 






and 



Igy :=fx^dF =Cx^ 2 7/dx. 



From the equation of the bounding curve 



a 



so that 



/„„ = — I x^ Va^ — x"^ dx 
a ^0 



MOMENT OF INERTIA 



119 



4^ 
6a% 



■-(2 x^ — a^)Va^ — x^.-\ sin ^ - 

8 8 a 



, and therefore, kgy = ^ 




Fig. 118 



In a similar way 

I ox = f,fdF = j f 2 xd7/ 

^0 



^^^, and therefore kgj.= a 



Since Ip = Ig^. -\- Igy, the polar moment of inertia with 
respect to the center is 



ab 



IT 



(a2 + h'^)^ and k^ = l^a^ + b\ 



It is seen that when a = b = r^ the equations obtained 
for the elliptical area are the same as those obtained for 
the circular area, just as they should be. 



120 



APPLIED MECHANICS FOR ENGINEERS 



73. Moment of Inertia of Angle Section. — When an area 
may be divided up into a number of triangles, or rec- 
tangles, or other simple divisions, the moment of inertia 
of the whole area with respect to any axis is often most 
easily found by taking the sum of the moments of inertia 
of the individual parts. Tliis method is often made use of 
in determininsr the moment of inertia of such areas as the 

o 

section of the angle iron, shown in Fig. 119. 

ffy 















II 
y' 


< — 




1 

1 y 


- \ 

■ \ 


J< 


> 


y 1 

1 




\ 




y 


1 

y 1 




' 






^ =" i 




^^ 


*^ y- 


^ 1 

1 






1 

1 







.9^ 



9^ 



Fig. 119 



We shall now determine the moment of inertia of this 
section with respect to the horizontal and vertical gravity 
axes, Igj. and Igy^ and also with respect to an axis v (see 
Art. 76), making an angle a with the axis x. Consider 



MOMENT OF INERTIA 



121 



the section divided into two rectangles, one 5" xf ", which 
we may call F^, and the other 3|" x |", which we may call 
jFg. The moment of inertia of the section, with respect to 
X, is equal to the moment of inertia of F^ with respect to x 
plus the moment of inertia of F^ with respect to a;, so that 

^,. = tV(5)(I)' + 5(f) (-808)2 + _,_(.2^I)3 I + .2^.(5) (1.19)2 

= 7.14 in. to the 4th power. Similarly 

I.. = tV(¥)(I)' + ¥(f)(l-31)^ + tV(|)(5)' + 6(|)(.88)2 
= 12.61 in. to the 4th power. 

Note. The problem of finding the moment of inertia of angle 
sections, channel sections, Z-bar sections, and the built-up sections 
shown in Figs. 121-125, is of special interest and importance to 
engineers, occurring as it does in the computation of the strength of 
all beams and columns made up of these shapes. 

Problem 124. Find 
the moment of inertia of 
the Z-bar section shown 
in Fig. 120 for the grav- 
ity axes gj. and gy. 

Hint. Divide the area 
into three rectangles. 

Problem 125. Com- 
pute the moment of 
inertia for the channel 

section, shown in Fig. 40, 1 Fig. 120 

Problem 31, for the horizontal and vertical gravity axes. 

Problem 126. Required the moment of inertia of the T-section 
(Fig. 41, Problem 32), also the moment of inertia of the U-section 
(Fig. 42, Problem 33) with respect to both horizontal and vertical 
gravity axes. 

Problem 127. The section shown in Fig. 121 consists of a web 
section and 1 angles, as shown. Find the moment of inertia of the 



-.gx 




122 



APPLIED MECHANICS FOR ENGINEERS 



whole section with respect to the horizontal gravity axis. Given, the 
moment of inertia of an angle section with respect to its own gravity 
axis, g', is 28.15 in. to the 4th power. 

Problem 128. Consider the section given in Problem 127 to be 
so taken that it includes two |-inch rivet holes, as indicated by the 



^ 



m 



m 



v" 






6"V- 



F='— 9' 



^ 



^ 



F^ 






J.- 



1.78 



y-^. 



h6 



d 



'—g' 



-14 






Fig. 122 



Fig. 121 

position of the rivets in Fig. 121. Compute the moment of inertia of 
the M^hole section, when the moment of inertia of the rivet holes is 
deducted. The distance from the center of the rivet hole to the out- 
side of the angle sec- 



'^ 



* T 






1.82 



± 



t 



3 



^-^^ — . I I tion may be taken as 

~-:~tTi M ^' -^ i"- Compare the 



result with that obtained in 
the succeeding problem. 



■<J 



^18 

Fig. 123. 



Problem 129. The same 
section shown in Fig. 121 is 
shown in Fig. 122 with two 
cover plates. Find the mo- 
ment of inertia of the whole beam section with 
respect to its horizontal gravity axis, now that 
the cover plates have been added. Allow for 
^-inch rivet holes. 

Problem 130. Find the moment of inertia 
of tlie section of a box girder, shown in Fig. 123, 
with respect to its horizontal gravity axis. 
The moment of inertia of one of the angje sec- 



MOMENT OF INERTIA 



123 



tions with respect to its own horizontal gravity axis, is 31.92 in. to 
the 4th power. 



XT 



i« 



-ee: 



v^ 



li ' 



\z^ 



d 



gx 



-16- 



9U 



Fig. 124 



Problem 131. Find the moment of inertia of the column section, 
shown in Fig. 124, with respect to the two gravity axes gx and gy. 



ffy 

i 



^ — t — f—^ 

i 'z^ — f 



I iy;6. 



nsyJ-L: 



< — 6 >, 



1.78" 
^x 



:rzTr__., 



gx 



-1-28- 



1 



Fig. 125 



124 



APPLIED MECHANICS FOR ENGINEERS 



The column is built up of one central plate, two outside plates, and 
four Z-l)ars, The legs of the Z-bars are equal, and have a length of 
3^ in. The moments of inertia of each Z-bar section with respect to 
its own horizontal and vertical gravity axis are 42.12 and 15.44 in. to 
the 4th power, respectively. 

Problem 132. Find the moment of inertia of the section shown 
in Fig. 12.5, with respect to the horizontal and vertical gravity axes 
gj. and g,j. This section is made up of plates and angles. The 
moment of inertia of each angle section with respect to both its own 
horizontal and vertical gravity axes is 28.15 in. to the 4th power. 

74. Moment of Inertia by Graphical Method. — It will 
often be necessary to find the moment of inertia of a plane 
section whose bounding curve is of a complicated form, as 




Fig. 126 



in the case when it is necessary to compute the strength 
of rails or deck beams. The graphical method given 
below may be used for such cases. 

Let F be the area bounded by the outer curve of Fig. 



MOMENT OF INERTIA 125 

126 and let OX he a line with respect to which it is desired 
to find the moment of inertia of F. 

Draw a line MJV parallel to OX at a convenient dis- 
tance, a, from OX. Let AB be any chord of the bound- 
ary curve of F parallel to OX. Project AB on MN, 
obtaining CD. From C and D draw lines to any point 
P on OX, cutting AB in A^ and B^ Project A^B^ on 
MJV, obtaining C^Dj. Join (7j and Z>j to P by lines 
cutting AB in A^ and B^. If this be done for all positions 
of AB, a new curve is obtained, as is shown in Fig. 126. 
Let F" be the area bounded by this curve. 

Letting AB = x, A^B^ = x', and ^2^2 = ^'^ ^^ ^^ seen 
from similar triangles that 

X a ^ x' a 
- = - and — =-, 
X y x y 



P 1*1 X Ct d 

irom which — - = — or x = — x 

x" y^ y 



Then, if J= moment of inertia of F with respect to OX, 

I=.^yixdy =j-a:^x"dy = a?^dF'\ 
or I=a;^F". 

The area F" may be measured by a planimeter or by 
one of the methods previously explained (Arts. 40-42). 

If it is desired to measure the area of F" by Simpson's 
Rule, the area F may first be divided into an even number 
of strips of the same width by horizontal lines and on the 
lines thus drawn the points of the auxiliary curve deter- 
mined by the method described above. The area F" 
will then be ready for the application of Simpson's Rule. 



126 APPLIED MECHANICS FOR ENGINEERS 

If, in laying off the area F^ 1 inch parallel to (9X repre- 
sents m inches and 1 inch perpendicular to OX represents 
n inches, then, measuring a in inches and F" in square 
inches, the value Jin inches^ is given by 

It is left for the student to show that the distance, ^, of 
the center of gravity of F from OX is given by 

_ aF' 

where F^ is the area of the curve traced out by A^ and B^ 
in Fig. 126. 

Problem 133. By the above method find the moment of inertia 
of the area of a circle about a diameter. Determine the area of F" 
by the use of Simpson's Rule. 

Problem 134. By the above method find the moment of inertia 
of a rectangle about one side. Take the point P at one of the vertices 
of the rectangle and show that the boundary curve of F" is a 
parabola. 

Problem 135. Find the moment of inertia of the area of the rail 
section of Fig. 68 with respect to the base line. 

Using the values of the area and height of the center of gravity 
found for the figure in Problem 72, find the moment of inertia of the 
area with respect to a horizontal gravity axis. 

75. Moment of Inertia of an Area, (a) by Direct Addition, 
and (b) by Use of Simpson's Rule. 

(a) Divide the area into n narrow strips of equal 
width Ax parallel to tlie line OY with respect to which 
the moment of inertia is desired (Fig. 127). If the 
middle lengths of these strips are respectively ^1,^2' '** ^n» 



MOMENT OF INERTIA 

and their distances from OY respectively Xy x^, 
the moment of inertia of the y 
area is given approximately 
by the formula, 

1= x\A^ + xlA^ + ••• + xlAr,, 
h — a 



127 



or, since A.r = 



, and. 



n 








Fig. 127 



approximately, A^ = y\^^^ 
A^ = y^/\x^ •" A^ = y^t^x, the approximate formula, be- 
comes 

h — a 



J = 



n 



[o-fi/j+o:-^?/, + ••• +a^;,y,J- 



Y 



A more accurate result is usually obtained by the use of 
Simpson's Rule. 

(5) Divide the area into an even number of strips of 

width Aa: parallel to the 
line with respect to which 
the moment of inertia is 
V,, sought (Fig. 128). 

Since / = I x^ydx^ where 

y is the variable length of 
>| the strips, it may be eval- 
uated by Simpson's Rule 
as in Art. 41, the quantity 



.'/o .'/i Ik % 



-Zi 




Fig. 128 



X 



hj taking the place of «/ in I ydx. Hence, approximately. 



jr = 



h — a 
3n 






128 APPLIED MECHANICS FOR ENGINEERS 

Problem 136. By the direct addition formula compute the mo- 
ment of inertia of a rectangle 4 in. by 2 in. with respect to a 2 in. side, 
using (1) 5 divisions, (2) 10 divisions. Compare with the exact value. 

Problem 137. Solve the preceding problem by the use of Simp- 
son's Rule, using (1) 6 divisions, (2) 10 divisions. 

Problem 138. Find the moment of inertia of the area of the sec- 
tion of Fig. 68 with respect to the base by the method of direct 
addition. 

Problem 139. Solve the preceding problem by the use of Simp- 
son's Rule. 

76. Least Moment of Inertia of Area. — In considering 
the strength of columns and struts it is necessary to know 
the axis about which the moment of inertia of a cross sec- 
tion is a minimum, since bending will take place about this 
axis. It was shown in Art. 65 that, if the moment of 
inertia of the area with respect to two rectangular axes in 
its plane fs known, the moment of inertia with respect to 
any other axis, making an angle a with one of these, could 
be found. It was further developed (Art. 67) that the 
value of a that would render the moment of inertia a min- 
imum was given by the equation 



tan 2 a = 



2 CxydF 
I. - 1. ' 



In case either of the axes x or ?/ is an axis of symmetry, the 
value of a given by this criterion is zero, so that, for areas 
having an axis of symmetry, the axis of least moment of 
inertia is the axis of symmetry or the one perpendicular to it. 
As an illustration of the problem in general let it be 
required to find the least moment of inertia of the angle 



MOMENT OF INERTIA 129 

section shown in Fig. 119 with respect to any axis in the 
plane of the area through the center of gravity. Let v be 
the gravity axis making an angle a with the .r-axis. The 
problem then is to find such a value of a that Ig^ will be a 
minimum. From Art. 65 we have 



Z 



gv 



= ly^ cos^ a — sin 2 a I xydxdy + Igy sin^ a. 



In Art. 73 it was found that I^^ ■= 7.14 and 7^^= 12.61. 

We proceed now to find the value of j xydxdy for the 
angle section. For this purpose, suppose the section com- 
posed of two rectangles, F^ (5 in. x f in.), and F^ (^^ in. 
X f in.), and then find the value of the integral for the 
two rectangles separately. Considering first the area F^^ 
and using the double integration, we get 

I I xydxdy = I xdx\ ^ — -^^ ^ 

Jl.62 J_.4'/5 ^ c/i.62 L 2 2 _ 

L 2 2 J 

In a similar way for F2, we have 

T' r-% 7 r-!5 r(2.88)2 (-.495)2- 

I I xydxdy = I xdxl ^ ^^ ^ 



.495 »^.995 L 2 2 



= 4 



.025[ai6 



62)2 (,995) ^ 



= 3.29. 



Therefore, | xydxdy for the whole area of the angle sec- 
tion is 5.51 in. to the 4th power. From this we find 



tan 2a = 11^ =2.02. 
5.47 



130 APPLIED MECHANICS FOR ENGINEERS 

Therefore 2 a = 63° 40', 

a=31°50'. 

The expression for Ig^ now becomes 

Ig, = 7.14 cos2 (31° 50')- 5.51 sin (63° 40') 

+ 12.61 sin2 (31° 50')= 3.72 in. to the 4th power. 
This gives the least radius of gyration, 

kg^ = .84 in. 

Problem 140. Find the least moment of inei-tia /^,, and least 
radius of gyration kg.,, of the Z-section shown in Fig. 120. In this case 
Ig^ = 15.44 in. to the 4th power and Igy = 42.12 in. to tlie 4th power. 

Anfi. Least /g-. = .5.61 in.- to 4th power and least kg^ = .80 in. 

Problem 141. An angle iron has equal legs. The section, similar 
to that in Fig. 119, is 8 in. x 8 in. with a thickness of I in. Find 
■igxj ■'■gyi isasi -ioD? anci least iCg^. 

Ans. Igx = Igy = 48.58 in. to the 4th power, 
Ig„ = 19.55 in. to the 4th power. 
kgy = 1.59 in. 

Problem 142. Find the moment of inertia of the column section, 
shown in Fig, 124, with respect to an axis v making an angle of 30° 
with gx. What value of a makes /^y mininmin in this case? 

77. The Ellipse of Inertia. — It is interesting to note, at 
this point, the relations between the moments of inertia 
with respect to all the lines, in the plane of the area, pass- 
ing through a point. We have seen that for every point 
in an area there is always a pair of rectangular axes for one 
of which the moment of inertia is a maximum and for the 
other a minimum; that is, there is alwa3^s a pair of prin- 
cipal axes. The criterion for such axes was found to be 

2^ 
tan 2 a = — zr, 

-£.,, "^ -'■•r 



MOMENT OF INERTIA 131 

which means, since the tangent of an angle may have any 
value from to infinity, positive and negative, that for 
every point there is always a pair of axes such that 



{ = 0, i.e. j xydF= 0. 



This means that the expression for Z„ may always be 
reduced to the form 

I^ = I J. cos^ a-{- ly sin^ a (Art. Q^) 

by properly selecting the axes of reference, where now 
Ij. and ly represent the principal moments of inertia. If 
we divide through by i^, the equation becomes 

^2 _ ^2 cos^ ct -\-kl sin^ «. 

k^ k^ ky 

,1 1 cos^a , sin^a 



or 



-J _ p2 cos^ at. .p^ sin^ a 



which is the equation of an ellipse referred to the princi- 
pal axes of inertia as axes, the coordinates of any point 
on the ellipse being x = p cos a, y = /o sin a. 

Hence if OX and OY are respectively the axes of 
maximum and minimum moments of inertia of an area for 
a given point 0, there exists an ellipse with minor and 
major axes on (9X and OF respectively such that the dis- 
tance from to the ellipse along any line is the reciprocal 
of the radius of gyration of the area with respect to that 



132 



APPLIED MECHANICS FOR ENGINEERS 




Fig. 129 



line. This ellipse is called the ellipse of inertia of the 

area for the given point (Fig. 129). 

The ellipse of inertia furnishes 
a graphical method of finding the 
moment of inertia of an area for 
any axis through a point when the 
principal moments of inertia of 
the area for that point are known. 

Problem 143. Sketch the ellipse of 
inertia of the area of a rectangle for (a) 
the intersection of the diagonals, (6) the 
middle point of one side of the rectangle. 

Problem 144. Determine the princi- 
pal moments of inertia of the area of 
a rectangle 6 inches by 10 inches for axes passing through a vertex 
of the rectangle. Sketch the ellipse of inertia. 

Problem 145. Determine the principal moments of inertia of the 
area of a circle for a point on its circumference, and sketch the ellipse 
of inertia of the area for that 
point. 

78. Moment of Inertia 
of a Thin Plate, (a) with 
Respect to an Axis in the 
Plate Parallel to its Faces, 
(b) with Respect to an Axis 
Perpendicular to its Faces. 
— (a) Suppose the plate 
to be of constant small thickness t and unit weight 7 
and let x be the distance of an element of mass c^iltf from 

the axis, where dM=^ ^tdF (Fig. 130). 




Fig. 130 



MOMENT OF INERTIA 



133 



Then approximately 

I=CxHM= ^-tSxHF, 

But j Tp'dF is the moment of inertia of the area of the 

face of the plate about the given axis. Therefore, ap- 
proximately, the moment of inertia of a thin plate with 
reference to an axis of the plate parallel to its faces equals 

— times the mometit of inertia of the area of its face with 

9 

reference to the same axis. 

(5) Let r be the distance of dM hom. the axis perpen- 
dicular to the face of the plate (Fig. 130). Then the 
moment of inertia of the plate with respect to the axis is 

I=CrHM= ^CrHF. 

Or, approximately, the moment of inertia of a thin plate 
with reference to an axis perpendicular to the faces of the 

ertia of the face of the plate with refer- 
ence to that axis. 



79. Moment of Inertia of Solid of Rev- 
olution. — Consider the moment of in- 
ertia of a solid of revolution with respect 
to its axis of revolution. Imagine the 
solid cut into thin slices, all of the same 
thickness, by parallel planes perpen- 
dicular to the axis of revolution (Fig. 131). Each slice 
is a circular disk of thickness dy and radius x., and its 




Fig. 131 



134 APPLIED MECHANICS FOR ENGINEERS 

moment of inertia with respect to the axis of revolution is 

-dyiraP' - — (Art. 71). The moment of inertia of the 
9 2 

solid of revolution is the sum of the moments of the small 

slices, so that 

the limits of integration and the relation between x and y 
depending upon the particular solid considered. 

Problem 146. Prove that the moment of inertia of a right cir- 
cular cylinder of radius r and altitude h with respect to its axis is 



Problem 147. Prove that the moment of inertia of a right cir- 
cular cone with respect to its axis is Ig^ = -^^ Mr^. 

Problem 148. Prove that the moment of inertia of a sphere 
with respect to a diameter is | A/r^. 

Problem 149. The surface of a spheroid is generated by re- 
volving the ellipse — + ^ = 1 about the x-axis. Prove that the mo- 
ment of inertia of the solid inclosed with respect to the axis of 
revolution is /^^ = | Mb"^. 

Problem 150. Prove that the moment of inertia of a rectangular 
parallelopiped with edges a, b, and c, with respect to a gravity line 

parallel to the edge c is Ig^ = ^— (a^ + b^). 

Problem 151, Prove that the moment of inertia of a slender rod 
of length a with respect to an axis perpendicular to the rod and 



passing (a) through the center, (h) through the end, is (a) 



12 
iVa2 



MOMENT OF INERTIA 



135 



Problem 152. An anchor ring is generated by revolving a circle 
of radius a about a line in its plane distant b from the center of the 
circle. Show that the moment of inertia of the mass of the anchor 
ring with respect to the axis of revolution is M{b- + | a-). 

80. Moment of Inertia of a Body by Parallel Sections. — 

By dividing a body up into thin plates by parallel planes, 
parallel to the axis with respect 
to which the moment of inertia 
of the body is sought, the mo- 
ment of inertia is made to de- 
pend upon the moments of 
inertia of the areas of the sec- 
tions and their distances from 
the given axis. 

As an illustration consider 
the problem of finding the mo- 
ment of inertia of a right circu- 
lar cone with respect to an axis through the vertex perpen- 
dicular to the axis of the cone (Fig. 132). The moment 
of inertia of the thin plate with respect to the diameter of 




its lower face is approximately 



71). The moment of inertia of this plate with respect to 
OX is therefore , . « 

^^-^-:- + - dy .y\ 

a 4: a 



y 



9 



or 



— f^ + ^yV?/, 



and the X for the cone is then 

9 



fir Jo V 4 



-^x^yAdy. 



136 



APPLIED MECHANICS FOR ENGINEERS 



From the figure x may be evaluated in terms of 1/ and 
the integral evaluated. 

Problem 153. Prove that the moment of inertia of a right cir- 
cular cone with respect to an axis through the vertex perpendicular 
to the axis of the cone is ^V M(r'^ + 4 A^). 

Problem 154. Using the result of Problem 153, find the moment 
of inertia of the cone about a gravity axis parallel to the base, and 

then about a diameter of the base, /o, = — Af ( r^ _) — ] . 

Problem 155. Prove that the moment of inertia of a right circu- 
lar cylinder with respect to a diameter of the base is Mi — f- — ) • 

Problem 156. Prove that the moment of inertia of an elliptical 
right cylinder of height h and semi-axes of base a and b with respect 

to the diameter 2 a of the base is All H — ), and with respect to a 

M 
gravity line parallel to the diameter 2 a is — (3 6^ + h^). 

Problem 157, Show that the moment of inertia of a right circu- 
lar cylinder, altitude h, and radius of base r, with respect to a gravity 

axis parallel to the base is Igx = ^^^(-7 H j, and find the moment 

of inertia with respect to an axis parallel to this and at a distance d 
from the base. 

Problem 158. It is required to find the moment of inertia of the 
cast-iron disk flywheel shown in Fig. 133 with respect to its geomet- 
rical axis. 







<^ 




^ 




# 



^ 



^ 

^ 



{^^^^^^^ 



Fia. 133 



MOMENT OF INERTIA 



137 



Hint. The wheel may be regarded as made up of three hollow 
cylinders, the moment of inertia of the whole wheel being equal to the 
sum of the moments of inertia of the three parts. The dimensions 
are as follows: diameter of wheel 2 ft., width of rim and hub 4 in., 
thickness of rim and web 2 in., thickness of hub 1^ in., and diam- 
eter of shaft 2 in. All distances must be in feet. 

Problem 159. Find the moment of inertia of the castriron fly- 
wheel shown in Fig. 134 with respect to its axis of rotation. There 
are six elliptical spokes, and these may be regarded as of the same 
cross section throughout their entire length. 




1 



■'::■■///:■ 



-.^--10 • > 






15 



Fig. 134 



Problem 160. Find the mass and moment of inertia, with respect 
to the axis, of a right circular cylinder whose density varies directly 
as the distance from the axis and is y\ at the outside of the cylinder. 



'^9 



5 



81. Moment of Inertia of a Mass ; Inclined Axis. — We 
shall now study tlie problem of tiiidiiig the moment of 
inertia of a solid with respect to an axis inclined to the 



138 



APPLIED MECHANICS FOR ENGINEERS 



coordinate axes. Suppose the moments of inertia of the 
body with respect to the three coordinate axes known 
from the expressions : 

and let it be required to find the moment of inertia of the 
body with respect to any other axis V making angles 

z 



dM 




Fig. 135 



a, /3, 7 with the coordinate axes. (See Fig. 135.) Let 
c?i^f equal the mass of an infinitesimal portion of the body 
and d its distance from the axis V. 

Since r^ = x^ + y"^ + z\ OA = x cos « + ?/ cos jB -{- z cos 7 
and 
d!^=zr^— OA^ = (aP' + y'^ -\- z^) — {x cos « + ^ cos ^ -\-z cos 7)^ 



MOMENT OF INERTIA 139 

we may write 
I^ = CdhlM 

= fli-^''^ + / + 2^) - i-^ cos rt 4- // COS yS + 2 COS yy]d3L 

This reduces, since cos^ « -f cos^ fi H- cos^ 7 = 1, to 
/„ = C(f -f- ^2) cos2 adM+fCz'^ + x2 ) cos2 /3(^il[f 

4- ) (a:2 + ?/2) cos2 ydM— 2 cos a cos ^ I xydM 
— 2 cos yS cos 7 I yzdM — 2 cos 7 cos a j xzdM^ 



or 



/„ = Jj. cos2 a-\- ly cos2 /3 4- ^2 cos2 7 — 2 cos a cos yS j xydM 
— 2 cos /S cos 7 j yzdM — 2 cos 7 cos « | xzdM^ 

which gives the moment of inertia of the body with 
respect to an inclined axis in terms of the moments of 
inertia with respect to the coordinate axes and the prod- 
ucts of inertia j xydM, | yzdM, and | xzdM. 

82. Principal Axes. — If the three products of inertia 
I xydM, I yzdM, and j xzdM are each equal to zero, the 
expression for I^ reduces to the form 

ly = Ij. cos2 a 4- ly cos2 /3 -{- I^ cos2 7. 

In this case the coordinate axes x, y, and z are called 
the principal axes for the point and the moments /,, I^, 
and I^, the principal moments of inertia for that point. 



140 APPLIED MECHANICS FOR ENGINEERS 

If the point is the center of gravity of the body, the 
principal axes are called the principal axes of the body. It 
can be shown that it is always possible to select the 
coordinate axes :r, ?/, and z so that the products of inertia 
given in the expression for /„ will each be zero. It fol- 
lows that for every point of a body there exists a set of 
rectangular axes that are principal axes. 

If the a:?/-plane is a plane of symmetry, the products of 

inertia j zydM^i\(\. | zxdM-dVQ both zero, for to any term in 

^zydM^ as z-^y-^dM^ there corresponds a term, — z^y^dM^ 
equal in numerical value but opposite in sign. Hence if 
two of the coordinate planes are planes of symmetry of a 
body, the three products of inertia with respect to these 
planes are zero, and the coordinate axes are the principal 
axes of the body through the origin. 

Problem 161. What are the principal axes of a sphere through 
a point on its surface ? What are the values of the principal mo- 
ments of inertia for that point ? | Mr^, I Mr% | Mr^. 

Problem 162. Find the moment of inertia of the ellipsoid whose 
surface is given by the equation 

-V.2 9/2 ^2 

a^ b'2 c^ 

with respect to the axes a, b, and c, and with respect to an inclined 
axis OV making angles a, (3, y with a, b, and c, respectively. The 
volume of an ellipsoid is 

3 5 o o 

and /„ = Ta cos- « + h cos^ jS -\- Ic cos*^ y. 

83. Ellipsoid of Inertia. — It is always possible to re- 
duce the expression for I^ to the form 



MOMENT OF INERTIA 141 

1^ = 1^ cos^ a. + ly cos^ P -{• Iz cos^ 7, 

by selecting the axes x^ y^ and z so that the products of 
inertia are zero. 

Dividing this equation througli by M^ we have 

kl = kl cos^ a -\- k'^y cos^ ^ + k^ cos^ 7. 

Let p = — , a = — , b = —^ and c = — . Then the equa- 
kv fCj. Ky kz 

tion becomes, on multiplying by p\ 

p^ cos^ a p^cos^ /3 /9^ cos^ 'y _ 1 

G? IP' C^ 

which is the equation of an ellipsoid of semi-axes a, 5, and 
c, on the principal axes, the coordinates of any point on the 
surface being 

a;=/3cosa, y = p cos /3, z — pco^f^. 

Hence for any body there exists for each point an 
ellipsoid such that the distance from the given point to 
the surface along any line is the reciprocal of the radius of 
gyration of the body with respect to that line, the axes of 
the ellipsoid coinciding witli the principal axes for that 
point. 

Since one of the semi-axes, a, 6, c, of the ellipsoid is a 
maximum value and another a minimum value of the dis- 
tance from the center to points on the surface, it follows 
that two of the principal moments of inertia of a body 
for any point are respectively minimum and maximum 
moments of inertia of the body with respect to lines 
through the point. 



142 



APPLIED MECHANICS FOR ENGINEERS 



Problem 163. Write the equation and construct the inertia ellij)- 
isoid for the contcr of gravity of a right circular cylinder, allitiidc h 

and radius r. 

Problem 164. Construct the 
inertia ellipsoid for the center 
of a solid .s]>here of radius r. 

Problem 165. Show that tlie 
moment of inertia of the seg- 
ment of the circle F\ (Fig. 136) 
with respect to the axis ()Z is 

the moment of inertia of the 
sector OBSA , minus \ ah^, the 
moment of inertia of the tri- 
angle OAB, or 

Fig. 136 7^^ := ^^^2 « - sin 2 a], 

and the moment of inertia of F■^^ with respect to OS is , ( q — ,-. sin a j, 
the moment of inertia of the sector, minus I ha^, the moment of inertia 

^4/4 I \ 

of the triangle A OB, or 7^^.. = -(«-., sin a + - sin 2 « . 

b^ O O / 

Problem 166. Show that the moment of inertia of the counter- 
balance (Fig. 58), with respect to a line through 0, perpendicular to 
05, and in the plane of the wheel, is approximately 

/,^=[^(2«-sin2«).-^;(2^-si„2^)+iiLip_F,(00')-^]^', 




= ^; - „r. 



w^here F^ = t^ — ai\ cos ^. 00' — r, cos ^ — r cos ?, and t is the thick- 
"9 ^ O ' o 2 

ness, as explained in Art. 38. 

Problem 167. Find the moment of inertia of the counterbalance 
(Fig. 58), with respect to a line through perpendicular to the plane 



MOMENT OF INERTIA 



U3 



of the wheel. It may be written 

Io= [|'(2 a - .^sin 2 a - ^sin «) - '^(2 /S - Urn 2 fi - |siii ^) 

7;(00')-]' 



+ 



^aWO' 



3 



84. Moment of Inertia of Locomotive Drive Wheel. — The 
drive wheel may be represented as in Fig. 137, and may be 
considered as made up of a tire, rim, twenty elliptical 
spokes, counterbalance, and equivalent weight on opposite 
side of center, and hub. 




Fig. 137 



The dimensions of the wheel are as follows : 
Tire, outside radius 40'^ inside radius 36", width 5". 
Rim, outside radius 36", inside radius 34", width 4J". 
Hub, outside radius 10^', inside radius 4|", thickness 8". 



144 APPLIED MECHANICS FOR ENGINEERS 

Counterbalance, outside radius 34", inside radius 
7'11.5'^ thickness 7|'^ 

20 spokes, 24" long, elliptical 3^'^ by 21". 

Angle at center subtended by counterbalance, «= 94° 40'. 

Radius of crank-pin circle 18". 

7 = 490 lb. per cu. ft. 

From the above data the following additional values are 
computed (Art. 38): 

Angle subtended by counterbalance at center of inner 
boundary circle, /3 = 30° 20'. 

Mass of counterbalance 17.1. 

Distance of center of gravity of counterbalance from 

center of wheel 28.8". 

28 8 
Mass of weights carried by crank-pin = ^^-^- x 17,1 = 27.4. 

18 

(Since moment of counterbalance = moment of crank- 
pin weights.) 

The moment of inertia of the wheel with respect to its 
axis of rotation is first found. The tire, rim, and hub are 
hollow cylinders, and their moments of inertia are com- 
puted from the formula 

For the counterbalance the formula of Problem 167 is used. 
The spokes are regarded as elliptical cylinders with the 
short axis of the ellipse in the plane of the wheel. Tlie 
moment of inertia of one spoke is computed from the 
formula 

TEhf^'U + ^^dx, (Arts. 77, 72, 64.) 



MOMENT OF INERTIA 145 

and 10 per cent is deducted for the parts of the spokes in- 
cluded in the counterbalance and boss. 

The moment of inertia of the weights at the crank-pin 
is computed on the assumption that the whole weight is 
concentrated at the center of the crank-pin. 

The results obtained are the following : 

Tire 7^ = 423 

Rim 136 

Hub 7 

Spokes 81 

Counterbalance 120 

Weights at crank-pin 62 

Total lo =S29 

With respect to a gravity axis OZ (Fig. 136) in the 
plane of the wheel, when the counterbalance is in a posi- 
tion where the line joining its center of gravity to the 
center is perpendicular to OZ, we get for the moments of 
inertia of the various parts : 

Tire I^^ = 212 

Rim 68 

Hub 4 

Spokes 41 

Counterbalance 103 

Weights at crank-pin .... 62 

Total /^^ = 490 

Since the widths of tire and rim are small compared to 
the diameter of the wheel, their moments of inertia are 
closely approximate to the values they would have if the 

L 



146 APPLIED MECHANICS FOR ENGINEERS 

material of tire and rim were in one plane, which would 
be one half of the corresponding Iq. The same is true, 
less accurately, of spokes and hub. The values of Iq have 
been divided by 2 for the corresponding values of Iq^ for 
the four parts mentioned. The weights at the crank-pin 
center are at the same distance from OZ as from 0, and 
the moment of inertia is therefore tlie same as for 0. The 
moment of inertia of the counterbalance was computed 
from the formula of Problem 166. 

Problem 168. Compute the moment of inertia of a pair of 
drivers and their axle with respect to their axis of rotation. Use the 
data given above and assume the axle as cylindrical, the diameter 
being 9}," and the length 68". Ans. 1661. 

Problem 169. Compute the moment of inertia of the pair of 
drivers and their axle, given in the preceding problem, with respect 
to an axis midway between the wheels and perpendicular to the axle. 
Consider the counterbalance of both wheels in such a position as to 
give a maximum moment of inertia and the distance between the 
centers of the wheels 60". 

Problem 170. Find the moment of inertia of two cast-iron car 
wheels and their connecting steel axle with respect to («) their axis 
of rotation, (h) an axis midway between the wheels and perpendicular 
to the axle. Consider the car wheels as composed of an outside 
tread, a circular web, and a hub ; each part may be considered a hollow 
cylinder with the following dimensions: tread, outside radius 16", 
inside radius 14", width ol" ; web, outside radius 14", inside radius 
.5^", thickness 1.5" ; hub, outside radius o^", inside radius 2\", width 
8" ; axle (considered cylindrical), 5" diameter and 7' 8" long. Dis- 
tance between centers of wheels 60". According to the assumption 
made above, the flange has been neglected, the web is considered a 
hollow disk, and the axle of uniform diameter throughout its 
length. 



MOMENT OF INERTIA 147 

Problem 171. The value 490 is the greatest vahie for the moment 
of inertia of a drive wheel with respect to a gravity axis in its plane. 
The least value will be with respe'ct to an axis at right angles to this 
through the centers of gravity of the counterbalance and wheel. The 
student should compute this least moment of inertia. 

Problem 172. In Problem 169 the drivers have been considered 
as having their cranks in the same plane. In practice they are 90^ 
apart. Find the moment of inertia with respect to the axis stated 
when the wheels are so placed. 



CHAPTER VIII 



FLEXIBLE CORDS. THE ARCH. BENDING MOMENTS 

85. Introduction. — A cord under tension due to any 
load may be considered as a rigid body. In the analysis 

of problems in which 
such cords are consid- 
ered, the method of 
cutting or section may 
be used. Since the cord 
is flexible (requiring no 
force to bend it), it is 
easy to see that, no 
matter what forces are 
acting upon it, it must 
T' have at any point the 
direction of the result- 
ant force at that point, 
and so must be under 
simple tension. If the 
cord is curved, as is the case where it is wrapped around 
a pulley, the resultant force is in the direction of the 
tangent. 

Consider, as the simplest case, a weight W suspended 
by a cord, as shown in Fig. 138 (a). The forces acting 
on TFare shown in (J) of the same figure. The cord has 
been considered cut and the force T^ acting vertically 

148 




FLEXIBLE CORDS 



149 



upward, has been used to represent the tension. Summa- 
tion of vertical forces = gives T = W. In Fig. 138 (c) 
the weight W^ is supported b}^ two cords. The system of 
forces acting on the point is shown in (c?), where T' 
and T' represent the tensions in the cords A and B 
respectively. ^X = and ^Y = give 

T' cos a = T" cos /3 
and r sin a + T" sin /3 = W^. 

These two equations are sufficient to determine the 
unknown tensions T' and T^'. 

Problem 173. A weight of 500 lb. is attached to the ends of 
two cords of length 8 ft. and 6 ft., the other ends of the cords being 
attached to the points A and B respectively, where .1 is lower than 
B, and is distant 9 ft. horizontally and 4 ft. vertically from B. Find 
the tensions in the cords. Solve analytically and graphically. 

SuGGESTiox. Make use of the horizontal and vertical projections 
of the cords to determine the angles which the cords make with the 
horizontal. 

Problem 174. A weight of 275 lb. is knotted at a point C to a 
rope which passes over two smooth pulleys at A and B distant 50 ft. 
apart and in the same hori- 
zontal line, carrying weights of 
350 lb. and 300 lb. respectively, 
as in Fig. 139. Find for what 
position of C the weights will 
be in equilibrium. Solve ana- 
lytically and graphically. 

Fig. 139 

Problem 175. If, in the pre- 
ceding problem, B is 10 ft. lower than A and distant 50 ft. horizon- 
tally from A, find the position of C for equilibrium. Solve analyti- 
cally and graphically. 




150 



APPLIED MECHANICS FOR ENGIXEERS 



mmmm. 
1) 




86. Several Suspended Weights. Analytical Method of 
Solution. — If two weights W^ and W^ are attached to the 

cord, as shown in 
the case of the cord 
ABQD (Fig. 140), 
each portion is un- 
der tension. Consider 
the cord cut at A 
and D and represent 
the tensions by T^ and Tg respectively. From2X=0 
and 21^= we have 

T-^ cos 7 = ^2 cos a 
and 2\ sin ^ ^ T^ sin « = W^ + ^Y^, 

A consideration of tlie forces acting at B^ if we call the 
tension in the portion BC^ T^^ gives, when tlie summation 
of the X and y forces are each put equal to zero, 

T^ cos 7 = 7^3 cos fi 
and 7\ sin 7 - T^^ sin /3 = ^¥^. 

In a similar way, consider the forces acting on the point 
(7, and we have 

T^ cos yS = 2^2 COS a 

and Tg sin /3 -}- 2^2 ^^^^ " = ^^v 



Of tlie six equations given above only four are independ- 
ent ; consequently, of the six quantities 7\, 2^2' ^3' "' A 
and 7, two must be known in order to determine the other 
four, or else additional conditions must be given for deter- 
mining two more independent equations. Such condi- 
tions may, for example, be the lengths of the cords and 



FLEXIBLE CORDS I'jl 

the positions of the two points of support, A and Z>. of 
iMg. 140. 

In general, if there are n knots such as B and C of 
Fig. 140, with the weights TFj, W^, TFg, TF^, etc., attached, 
it will be possible to get n + 2 independent equations. 
Tliese will be sufficient to determine the tension in eai-h 
portion of the cord and its direction, provided the tension 
at A^ say, and its direction are known. If the weights 
are close together, tlie curve takes more nearly the form 
of a smooth curve. Two special cases of this kind are 
discussed in this chapter in Art. 97 and Art. 100. 

Problem 176. The points .1 and D of Fig. 140 are distant 20 ft., 
apart, and are in the same horizontal hne. Eacli of the cords is 
10 ft. long and the weights are each 50 lb. Find the tensions in the 
cords. 

Problem 177. In Fig. UO W^ = 50 lb., W., = 120 lb., AB = 10 ft., 
y =: (jO*^. J) and A are on the same level, 20 ft. apart. Find the 
tensions in CD and BC and tlie lengths of tliese cords. 

87. Graphical Method of Solution. — In Fig. 141 (z) is 
shown a cord carrying the weights IFj, TF^, TFg, and in 
(z7) the corresponding stresses in the cord and the re- 
action at the supports. The cord in this position may be 
thought of as a rigid body in equilibrium under the action 
of the weights and the reactions of the supports. The 
weights and the reactions of tlie supports therefore form 
a system of forces in equilibrium and the relation found in 
Art. 57 must exist here between the rays to the force 
polygon and the sides of the equilibrium polygon. 

In stating the problem for solution enough conditions 
must be given to determine the construction. If the 



152 



APPLIED MECIIAMCS FOR ENGINEERS 



points of support are given and assigned weights are to 
act in assigned vertical lines, then there exist an indefinite 
number of solutions. For the tensions in tlie strings can 
be increased or decreased by shortening or lengthening 
them. 

If the string were cut at any point and held in position, 
the part of the string to one side of the cut would be in 




ii) 



Fig. 141 



equilibrium under the action of the forces acting upon 
that part. The horizontal component of the tension at 
the point w^here the cut is made must then be equal to 
the horizontal component of the reaction at either support. 
The horizontal component of the tension of the string is 
therefore the same at all points and is equal to the hori- 
zontal component of either reaction. 

If, now, in addition to having given the points of sup- 
port, the weights and their lines of action, there is also 
given the horizontal component of the tension in the 
string, the graphical construction for the shape of the 



FLEXIBLE CORDS 153 

string and the tensions in the segments can be carried out. 
The force polygon agf"-ca is laid off, the point h not 
being determined. Any point is chosen and the rays 
oa^ og^ ••• oc are drawn. Horizontal and vertical lines 
are drawn through the points of support as action lines of 
the horizontal and vertical components of the reactions at 
these points. Beginning at any convenient point, as 
on AB^ the equilibrium polygon is drawn with sides 
respectively parallel to the rays of the force polygon. 
Parallel to the closing side (dotted) of the equilibrium 
polygon, a ray (dotted) is drawn from to ca determin- 
ing the point h. Therefore ch and ha are respectively the 
vertical components of the reactions of the right and left 
supports with the given value of the horizontal compo- 
nent of the tension. The triangle hag is then the triangle of 
forces acting at the left support, and hg therefore repre- 
sents the tension in the segment of the cord attached there. 
The segment of the cord BG may then be drawn parallel 
to hg from the point of support to the vertical line in which 
W^ acts. In the same way the triangles hgf^ hfe^ and hed 
are the force triangles for the points where the weights 
are attached and hf^ he, and hd represent the tensions in 
the segments BF, BE, and BD respectively. They give 
therefore the direction of these segments. The shape of 
the cord can then be constructed. 

A check on the accuracy will be that the last segment 
must pass through the right support. 

The segments of the cord form an equilibrium polygon, 
ov string polygon, for which the corresponding ray poly- 
gon has h as a pole. 



15-t APPLIED MECHANICS FOR ENGINEERS 

Problem 178. Weights of 25, 40, and oO lb, are to he suspended 
by a cord in lines distant 4, 8, and 11 ft. respectively from the left 
support. The right support is to be distant 15 ft. horizontally and 
3 ft. higher than the left. With a horizontal comi)onent of tension 
in the cord equal to 30 lb., find by graphical methods the shape of 
tiie cord, the lengths of the segments, the tensions in the segments, 
and the reactions at the support. 

Problem 179. Ten equal weights are to be hung at equal hori- 
zontal distances on a cord supported at two points on the same level. 
The horizontal tension in the cord is to equal five times the value 
of one weight. Draw the shape of the string and determine the 
maximum tension. Draw on the same figure tlie shape of the cord 
when the horizontal tension is three times the value of one weight. 

88. Locus of the Pole of the String Polygon. — Let the left- 
and right-hand vertical components of the reactions at 
the supports be Y^ and Y2 respectively, II the horizontal 
component of the tension, a^, a^-, a^ the horizontal distances 
of the weights W^^ TF^, W^ from the left support, I the 
horizontal distance between the supports, and h the verti- 
cal height of the right support above the left. 

Taking moments about the left support, 

Let y/^ ^^1^1 + ^2^^2 + ^3^3 

^ I 

Then F2 = Yl + -R. 

The value of P^ ^^ the value that Pg ^^'ould have if either 
h ov H were zero, i.e. if the supports were on the same 
level, or if the string were replaced b}" a rigid body simply- 
resting on the supports without horizontal pressure. 



FLEXIBLE CORDS 



155 



The value Y'^ may be computed analytically and laid off 
vertically from d on the load line gd (Fig. 142), or it may 
be located by using the equilibrium polygon for vertical 





forces only, assuming H to be zero. Y^ is then found by 
increasing Y'^ by the quantity - H^ which is proportional 

to H. If through the end, 6', of Y'^ a line is drawn 
parallel to the line joining the points of support, the pole, 
^, of the string polygon is found on this line at a horizon- 
tal distance H from the load line gd. Conversely, any 
[)oint on this line may be taken as a pole for a string 
polygon, and the corresponding value of H is the horizon- 
tal distance from the point chosen to the load line gd. 
The locus of the poles of the string polygons is therefore 
the straight line parallel to the line joining the points of 
support and passing through the points which would 
divide the load line into the reactions if the supports were 
on the same level. 



156 



APPLIED MECHANICS FOR ENGINEERS 



By varying the position of the pole the form of the 
string polj^gon may be changed. 

89. Pole Distance and Depth of String Polygon. — The depth 
of the string polygon^ measured from the line joining the 
'points of support^ varies inversely as the pole distance from 
the load line. For, if d is the depth of the string polygon 





at a certain point and D the distance of the pole b from 
the load line (Fig. 143), we have from similar triangles, 

(d + zy.x:: F/ :i), 
and z: (^x— a-^) : : W^i D. 

Eliminating 2, 

Since aj, Y\^ and Wi are constants, and for the given 
position X is constant, this equation shows that d varies 
inversely as D. The proof can be given in like manner 
for any point. 

The depth of the string polygon at any point can then 
be made to have any desired value by a proper choice of 



FLEXIBLE CORDS 157 

the pole. If with a pole distance I) the depth of the 
string polygon at a certain point is d, and a depth d^ is 
desired, choose a new pole distant D^ from the load line 

so that i)j = — , and the string polygon constructed by 

the use of this pole will be the one sought. 

Problem 180. Weights of 100, 300, 200 lb. are to be suspended 
from a cord in lines 3, 5, and 8 ft. respectively from the left support. 
The right support is distant 10 ft. horizontally and 2 ft. vertically 
above the left support. Using a scale of 1 in. = 2 ft., construct a 
string polygon whose depth shall be 4 ft. at the second load. From 
the diagram scale off the tensions in each part of the cord and the 
horizontal component of the stress in the cord. Use a force scale of 
1 in. = 100 lb. 

Problem 181. A cord is suspended from tvs^o points on the same 
level 9 ft. apart. Eight weights of 50 lb. each are to be suspended at 
equal intervals between the supports. The maximum tension in the 
cord is to be 300 lb. Draw the shape of the cord. Find the total 
length of the cord and the horizontal component of the tension. 

Problem 182. Show that when equal weights are distributed at 
equal horizontal intervals along a cord suspended between two 
supports on the same level, the maximum tension in the cord is 
greater than half the total load on the cord. 

Problem 183. A cord supported at the ends on the same level 
carries a load uniformly distributed along the horizontal. Draw 
approximately the shape of the cord, given that the tension of the 
cord at the lowest point is one half the total load. 

Suggestion. Divide the horizontal distance up into small equal 
intervals and assume that the weight in each interval acts at the 
center of that interval. 

With the vertex at the lowest point of the string polygon, construct 
a parabola to pass through the points of support and compare it witli 
the string polygon. 



158 



APPLIED MECHANICS FOR ENGINEERS 



Problem 184. Make the constructions of the above problem when 
the points of support are not on the same level. 

Problem 185. Show how to obtain the ratios of a set of weights 
to cause a cord to liang in a given string polygon. 

Hint. Work from the string polygon to the force polygon. 

90. The Linked Arch. — Analogous to the problem of 
finding the form taken by a cord carrying loads at inter- 
vals is that of finding the form of a set of connected 
weightless links to sustain a given set of loads in given 
vertical lines, the links to be in compression instead of 
tension. Here again if the horizontal component of the 





Fig. 144 



thrust in any link is given, the form of the linkage can be 
determined. The construction and proof are analogous to 
those for the cord carrying weights. (See Fig. 144.) The 
links themselves lie in an equilibrium polygon the pole of 
which is b. It will be noted that the pole of the link 
polygon lies on the opposite side of the load line from the 
pole of the string polygon, and that all poles of link 
polygons lie on the same straight line as that on which 
the poles of the string polygons lie. 



FLEXIBLE CORDS 



159 



91. The Masonry Voussoir Arch. — In an arch made of 
links, constructed to carry a given set of loads, any varia- 
tion of the loads, unless they were all changed simulta- 
neously in the same ratio, would be apt to cause the 
collapse of the arch. In an arch built of dressed stone, 
called voussoirs, owing to the size of the bearing surfaces 
and the friction between the surfaces, an arch can be built 
to carry a given set of loads and allow for a given varia- 



-\ — t 


r 7 — 


\ 




\ 




\ ' 












■^' \ > 


c / J^ 


\ ] 




\ I 




\ ^ 




Vl 




— t? 


Le-i 



-^ ^ 


< 7- 


\ ' 


' / 


\ I 


\ / 


\ ^ 




fl \ 5 


tiJ R 


Vi 


W 


V-t, 


1/ 


^ 


1 




(a) 



Fig. 145 



(c) 



tion in the loads without endangering the structure. 
The voussoirs can be made of such a depth as to keep the 
line of the resultant pressure of two adjacent voussoirs 
on each other in a certain portion of the joint. The 
broken liiie joining the points of the resultant pressures 
of the adjacent arch stones on each otlier is called the line 
of resistance. It is usually considered advisable, though 
it is not necessary for the stability of the arch, to keep the 
line of resistance within the middle third of the arch. The 
intensity of the pressure between two arch stones is 
assumed to vary uniformly as indicated in Fig. 145. In 
(a) the resultant pressure falls within the middle third, 
in (5) at the edge of the middle third, and in (c) outside 
the middle third. In the latter case there is a tendency 



160 APPLIED MECHANICS FOR ENGINEERS 

for the joint to open. With the line of resistance below 
the middle third there would be a tendency to open the 
joint on the upper part of the arch, and also to unduly in- 
crease the pressure on the lower part of the joint. vVhere 
the line of resistance falls above the middle third the joint 
would tend to open on the lower side. 

It is desired here to present only the elementary prin- 
ciple of the arch. An adequate treatment of arches may 
be found in I. O. Baker's " Treatise on Masonry" (John 
Wiley & Sons). 

Problem 186. Find the shape of a linkage that would carry the 
weights of Problem 178 with the same horizontal stress, the links 
being in compression. Is the link polygon the same as the inverted 
string polygon of that problem ? 

Problem 187. Show that when, and only when, the supports are 
on the same level, the link polygon for a given set of loads and 
horizontal stress is the inverted string polygon for that set of loads. 
Show also that in any case the link polygon can be obtained from the 
string polygon turned through ISO"" in its plane if in constructing the 
string polygon the loads and their horizontal distances are reversed in 
order from right to left. 

Problem 188. An arch is to be constructed with its center line 
in the form of an arc of a circle, the span being 20 ft., and rising 6 ft. 
in the middle. Considering the .load on the arch to be vertical only 
and at any point proportional to the distance from the center line of 
the arch to a straight horizontal line 6 ft. above the highest point of 
the center line, sketch approximately the line of resistance. About 
how deep would the arch stones need to be to keep the line of resist- 
ance within the middle third ? 

Suggestion. Assume the loads concentrated at equal horizontal 
intervals and construct the link polygon with the required height of 
6 ft. It may then be seen how the line of resistance may be changed 



FLEXIBLE COEDS 



161 



so as to more nearly coincide with the given circle by a slight move- 
ment of the pole. 

Problem 189. Find approximately by graphical methods the form 
of an arch to carry a load uniformly distributed along the horizontal. 

92. Bending Moments. — As an application of the equi- 
librium polygon for parallel forces a brief discussion is 
here given of bending moments in a horizontal beam sub- 
jected to vertical forces only. Let Fig. 146 represent a 
horizontal beam acted on 
by vertical forces. At any p^ 
point C pass a vertical 
plane perpendicular to the 
axis of the beam. 

The sum of the moments 
about a horizontal line in 

this plane of all the forces acting on the beam to the left 
of the plane is called the bending moment at that section. 
Clockwise direction will be counted as positive. Repre- 
senting the bending moment by M^ the value of iltf at 6' 
(Fig. 146) is 



-aj-^ 



W LBS /ft. 






I. 



Fig. 146 



M= R^x -{x + a^)P^ - 



W(x - ay 



Since the sum of the moments of all the forces acting on 
the beam is zero, the sum of the moments of the forces to 
the right of the section is the negative of the sum of the 
moments to the left of the section. Hence, in comput- 
ing -bending moments, if it is more convenient, we may 
take as the bending moment at the section the sum of the 
moments of all forces acting: on the beam to the rig-ht of 
the section and consider counter-clockwise as positive. 



162 



APPLIED MECHANICS FOR ENGINEERS 



Problem 190. A beam 10 ft. long is supported at the ends and 
carries loads of 500 lb., 600 lb., 800 lb. at distances of 3 ft., 5 ft., and 
8 ft. respectively from the left end. Compute the bending moment 
at intervals of 1 ft. along the beam, neglecting the weight of the 
beam. 

Problem 191. A beam 12 ft. long, supported at the ends, carries 
a load of 500 lb. at a point 4 ft. from the left end and a uniformly 
distributed load of 200 lb. per linear foot over the right half of the 
beam. The beam weighs -30 lb. per foot of length. Compute the 
bending moment at intervals of 2 ft. along the beam. Using M as 
ordinate and distance along the beam as abscissa, sketch a curve repre- 
senting the bending moment at all sections. 

93. Bending Moment Diagram. — Using the bending mo- 
ment as ordinate and distance along the beam as abscissa, 
a curve may be plotted whicli shows the bending moment 




Fig. 147 



at each section. This curve is called the bending moment 
diagram. It will now be shown that for concentrated loads 
the equilibrium polygon represents to a certain scale, the 
bending moment diagram. 

Consider a beam supported at the ends carrying loads 
Pj, P^, Pg lb. at distances of Xp x^, j^^ ft. from the left end 



FLEXIBLE CORDS 163 

respectively. Construct the equilibrium polygon for the 
forces and reactions, using the following scale : 

V =m ft. along the beam, 
1" = n lb. on the line of loads. 

Let D be the pole distance in inches, and d the depth in 

inches of the equilibrium polygon at a point distant — in. 

m 

from the left support between the loads Pj and P^. 
By similar triangles 

m n 

and z : — ^I^i = — i : i). 

m 71 

Eliminating z, d = — - — ^ — ^^^^^ — - • 

° mnl) 

By definition the bending moment, M, measured in pound- 
feet at the given point is 

M=xR^-{x-x^)F^. 

,'. d = -, 

mnD 

or M—mnD'd. 

This formula can as easily be shown to hold for any other 
point. Hence the depth of the equilibrium polygon in inches 
multiplied by the number of ft. per inch along the beam^ the 
number of lb. j^er inch on the load line, and the pole distance 
in inches gives the value of the bending moment in lb. ft. 

Tlie depth of tlie equilibrium polygon then represents 
the bending moment to the scale 

1" = mnl) \h.-it. 



164 



APPLIED MECHANICS FOR ENGINEERS 



If it is desired to use the pound-inch as the unit of moment, 
then let 1" =m in. along the beam. 

To obtain the bending moment diagram to a given scale, 
1" = k lb. -ft., it is only necessary to choose D so that 
mnD = k. 

For a distributed load the bending moment curve may 
be obtained approximately by dividing the beam length 
into small intervals and considering the load on each in- 
terval to act at the center of that interval. 

Problem 192. Construct graphically the bending moment for the 
beam of Problem 190. Let 1 in. = 2 ft. along the beam, 1 in. = 500 lb. 
on the load line, and choose D so that 1 in. on the moment diagram 
shall equal 1500 Ib.-ft. of moment. 

Problem 193. Construct approximately the bending moment 
diagTam for a beam supported at the ends and carrying a uniformly 
distributed load. 

94. Bending Moment for Beams not Supported at the Ends. 
— (a) Cantilever Beam. A cantilever beam is shown in 
Fig. 148. The construction for the bending moment is 
indicated, the scale factors being the same as in Art. 93. 






P. 


1 




4 




1 


iT^ 


2\^ 


^ 






3\ 




J. 148 









FLEXIBLE COBDS 



165 



(J) Overhanging Beam. In Fig. 140 the construction 
is indicated for the bending moment of an overhanging 
beam. The diagram (i) is the equilibrium polygon. 
The diagram (ii) is obtained from (i) by replacing the 
closing line of the polygon and the two lines connected 





Fig. 149 

with it by a horizontal line and drawing a polj^gon with a 
depth on each load line and reaction line, the same as that 
of the equilibrium polygon {i). 

Problem 194. Prove that the metliod indicated of drawing the 
bending" moment diagram of a cantilever beam is correct. 

Problem 195. Prove that the method indicated of constructing 
the bending moment diagram of the overhanging beam is correct. 

Problem 196. Construct the bending moment diagram of a 
cantilever beam 12 ft. long carrying loads of 400, 600, and 1000 lb. 
at distances of 0, 3, and 7 ft. respectively from the free end. Scale 
the bending moment at 3 ft. and at ft. from the free end and com- 
pare with computed values. State the scales used. 



166 



APPLIED MECHANICS FOR ENGINEERS 



Problem 197. A beam 16 ft. long is supported at two pothts dis- 
tant respectively 3 ft. from the left end and 4 ft. from the right end. 
Loads of 300, 500, 1000, and 600 lb. are applied at distances of 0, 6, 
10, and 16 ft. respectively from the left end. Construct graphically 
the bending moment diagram to a convenient scale. Check by com- 
puting analytically the bending moment at several points. 



^^ 



A 



/\ 




R. 



Fig. 150 



95. Tensile and Compressive Stresses in Beams. — Let Fig. 
150 represent the portion of the beam of Fig. 146 to the 
left of the plane section at C. This portion of the beam 

is in equilibrium under the action 
of the external forces acting on 
this portion and the internal forces 
exerted on it by the portion of the 
beam to the right of the section. 
These internal forces can be re- 
solved into horizontal and vertical 
forces. The vertical component of 
the internal forces must be equal and opposite to the sum 
of the external forces to the left of the section. This 
latter sum is called the shear at the section. 

Since the external forces are all vertical, the sum of the 
horizontal components of the internal forces must be zero. 
The horizontal forces therefore form a couple. This 
couple is called the internal stress couple. To produce 
equilibrium the moment of the internal stress couple must 
be equal and opposite in sign to the moment about any 
horizontal line in the section of all the external forces to 
the left of the section. That is, the internal stress couple 
at any section of the beam is equal to the bending moment 
at that section changed in sign. 



FLEXIBLE COIiDS 



107 



It can be shown by experiment that when a rod is stretched 
or compressed within tlie ehistic limit, the amount of 
elongation or compression is directly proportional to the 
applied force. Defining unit stress in a rod uniformly 
stretched or compressed, i.e. without bending, as the total 
force divided by the cross-sectional area, and unit strain 
as the amount of elongation or compression divided by the 
original length, this experimental fact may be expressed 
by the equation. 



unit stress 



= 11 



unit strain 

where -E'is constant for a given material. 

^ is called the modulus of elasticity. Experiment shows 
that it is the same for compression as for tension, for such 
materials as wood and structural steel. 

It is also shown by experiment that when a beam is 
bent, within the elastic limit, a plane section of the beam 
before bending remains a plane section after bending. 
When the beam is bent, the upper fibers of the beam are 
then compressed (or elongated) and the lower fibers are 
elongated (or compressed) as in Fig. 151. There will be 



A' B' 




\ ^ — 1 

V- ^ 


1 7 t 


* «• / 1 


y ''1 




^ 


4.:....-L 


— i^ 



B 



Fig. 151 



Iti8 APPLIED MECHANICS FOR ENGINEERS 

somewhere between top and bottom a surface, called the 
neutral surface^ where there is neither tension nor com- 
pression, and the amount of tensile or compressive stress at 
any point in the section will vary directly as the distance 
of this point from the neutral surface, since the amount of 
the elongation or compression of the longitudinal fiber of 
the beam through the point varies as the distance from 
the neutral surface. 

If s is the unit compressive stress at the upper outside 
fiber of the beam, distant y from the neutral surface, s' the 
unit stress at a distance y' from the neutral surface, then 

y 

s' will be positive or negative according to the sign of y' . 
Considering the portion of the beam as in equilibrium 
under the action of the external and internal forces, if dA 
is an element of area of the cross section distant y' from 
the neutral surface, we have from 2X= 0, 

Cs'dA = 0, 

or "-Cy'dA^O. 

y^ 

But i y'dA is equal to the product of the area of the cross 
section and the ordinate of the center of gravity measured 
from the origin of ordinates, i,e. from the neutral surface. 
If y is this ordinate, then 

^yA=^. 

y 

Therefore y =^i 



FLEXIBLE CORDS 1G9 

and the neutral surface passes through the center of 
gravity of the section.* 

Equating moments of external and internal forces about 
a horizontal gravity axis of the section, 

Mom of external forces = — mom of internal forces, 

or M= Cy's'dA 

= -Cy'HA = -I 

y^ y 

where J is the moment of inertia of the area of the cross 
section with respect to the horizontal gravity axis of the 
section. 

This formula gives the stress s at a distance y from the 
neutral axis of the section. The stress at any other point 
in the section is obtained by the direct proportion 

s y' 

For a beam of constant cross section y and J are constant 
and hence 8 varies directly as M. The maximum tensile 
or compressive stress will therefore be found where the 
bending moment has maximum numerical values. 

In computing stresses care must be used to have the 
same units throughout the formula. If stress is reckoned 
in pounds per square inch, the dimensions of the cross 

* In the above it is assumed tliat the intersection of the vertical section 
and the neutral surface is a liorizontal straight line. If the loads lie in 
one plane which is a plane of symmetry of the beam, this will be the case. 
Under other conditions the neutral surface may not be liorizontal. 



170 APPLIED MECHANICS FOR ENGINEERS 

section must be in inches and the bending moment in 
pounds-inches. 

Problem 198. A beam 4 in. broad by 6 in. deep, 12 ft. long, is 
supported at the ends and carries loads of 400 lb. and 800 lb. at dis- 
tances of 5 ft. and 9 ft. respectively from the left end. Sketch the 
bending moment diagram and find the maximum fiber stress. 

Aris. 11.50 Ib.-sq. in. under the 800 lb. load. 

Problem 199. A beam 4 in. broad by in. deep, 12 ft. long, is 
supported at the ends and carries two equal loads at distances of 4 ft. 
from the ends. What maximum value may the loads have if the 
maximum allowable fiber stress is 1000 lb. per sq. in. ? 

Problem 200. Find the ratio of the loads that could be put on a 
beam 4" by 6" when placed with the 4" face horizontal and when 
placed with the 6" face horizontal, the maximum fiber stress to be 
the same in the two cases. 

Problem 201. Compare the loads, one concentrated at the middle, 
the other uniformly distributed over the length of the beam, to cause 
the same maximum fiber stress in the beam. 

Problem 202. An I-beam of depth 10 in. and weight 25 lb. per 
linear foot is 16 ft. long and is supported at the ends. What con- 
centrated load can it carry in the middle so as to cause a maximum 
fiber stress of 16,000 lb. per sq. in.? The moment of inertia of the 
cross section about a horizontal gravity axis is 122 in.^. Take the 
weight of the beam into consideration. 

96. Cords and Pulleys. — When a cord passes over a 
pulley, without friction, the tension is transmitted along 
its length undiminished. A weight W attached to a cord 
which passes vertically over a pulley is raised by a direct 
downward pull P on the other end of tlie rope. If there 
is no friction, P is equal to TT, for uniform motion. In 
the case of a system of pulleys, as shown in Fig. 152, the 



FLEXIBLE COEDS 



171 



cord may be considered as under the same tension through- 
out and parallel to itself in passing from one sheave to tlie 
other. It is then possible to cut across 
the cords, just as was done in the case of 
the bridge truss, Problem 92, where the 
stress was along the member in each case. 
Cutting all the cords at C and considering 
all the forces acting on the sheave B^ we 
get, calling the tension in the cord P, 

or the tension in the cord is W/Q. A con- 
sideration of the upper sheaves gives 
r= 7 P = 7/6 ( TF). The various cases 
of cords and pulleys that come up in 
engineering work may be taken up in a 
similar way, but in any case of cutting 
cords, it must be remembered that all 
cords attaching one part to another must 
be cut and the tension acting along the cords inserted be- 
fore the principles of equilibrium can be applied. The 
consideration of the friction between cords and pulleys will 
be taken up in Chapter XIII. 

Problem 203. Suppose the hook of the lower sheave of Fig. 152 
attached to a weight of i]50 lb., and a man of weight 150 lb. stands 
on the weight and lifts hinjself and the weight by pulling on the rope. 
What force must he exert? If he stands on the ground, what force 
must he exert to raise a weight of 500 lb. ? 

97. Equilibrium of a Flexible Cord Carrying- a Load Uniformly 
Distributed along the Horizontal. — Let the cord supported at 




Fig. 152 



172 



APPLIED MECHANICS FOR ENGINEERS 



A and B (Fig. 153 (<z) and (6)) carry a uniform load w per 
unit length along the horizontal. Consider a small portion 
of the cord, As, with ends at (a;, ^) and (a: + Ax, y + A^), 
the coordinate axes being horizontal and vertical. This 

Y 





(a) 



(6) 
Fig. 153 



(c) 



portion is in equilibrium under the action of the tensions T 
and T^ at its ends and the load wAx carried by this portion 
of the cord (Fig. 153 ((?)). 

Resolving T and T^ into horizontal and vertical compo- 
nents and appl3dng the conditions for equilibrium, we get 



But 



X'-X=0, 

r' _ Y= wAx. 

Y' — Y= increment of Y = AY. 



(1) 

(2) 



Equation (1) shows that the horizontal component of the 
tension is the same throughout the cord, and equation (2) 
may be written 

dY 

ax 

Y= wx + c. 



and hence, 



(3) 



Now X and Y are respectively equal to Tcos6 and T 
sin 0, where 6 is the inclination of the tangent line to the 



FLEXIBLE CORDS 173 

curve at (x, ?/). Equations (1) and (3) may then be 

written 

Tcos 6 = X, 

Tsin 6 = wx -\- c. 

By division we ofet, since tan 6 = ^, 
-^ "" dx 

dy _wx -{- c ^ . >^ 

Ix ~ X ^ ^ 

Since c and iv are constants, a value of x can be found for 
a point on the curve, not necessarily on the cord -between 

A and J5, for which -^ = 0. If the origin be moved to this 

dx 

point on the curve by a translation of axes, equation (4) 
will take the form 

^ = — CS^ 

dx X 

This means that the origin is taken at the lowest point of 
the curve as in Fig. 153 (a) and (6). In (a) the origin 
is on the cord, and in (J) it is not. 
Integrating equation (5), 

y^^^' + o'. 

But y = when x = 0. Therefore c' = and the equation 
of the curve becomes 

^ 2X 

Hence the cord hangs in a portion of a parabola with ver- 
tical axis. 

98. The Horizontal Component of the Tension in Terms of 
the Length of Span and the Deflection. 



174 APPLIED MECHANICS FOR ENGINEERS 

The horizontal distance between the supports is called 
the span^ I. 

The vertical distance of the vertex of the parabola 
below the lower point of support is called the deflection, d. 

In Fig. 153 (a) and (J) the coordinates of A are 
( — Zj, c?) and (Zj, d) respectively, and the coordinates of 
B are (Zg, d ■]- h'). The coordinates of A and B satisfy 
the equation of the curve. Hence in case (a) 

cl=:J^l2 and d-\-h = -^Il 
Therefore, ?, = J^^, Z. = J^^(Z±5. 



Therefore Z = /^ + /, = V".~ ("^^ + ^'^ + ^)' 
and X= 



w 



If h = 0, the supports are on the same level and 



In case (6), 1 = 1^ — \ = ^" (c? + A) — -y^*" -, 

and -X' = 






2(Vd-\-h-vdy 

99. Length of Cord. — The length of the curve from the 
origin to any point (a:, ?/) on tlie curve is given by 

/- 



-XV+fST-^- 



FLEXIBLE CORDS 175 

Letting — = a, the equation of the parabola becomes 
zv 

or • ft 11 ^ 

y = - — , from which -f- = -, and hence 
2 a ax a 

^ 2a 



\^a^ + l'i-\-ano^. 



I a 



?2 V d^ ^l\-\- a^ log, 



h 


+ Va2 


+ l'\ 




a 




h 


+ Va2 


+ li 



a 

The total length of the curve is 

in case (a) S = 8-^-\- %^^ and 
in case (6) *S'= 8^ — s^. 

Another formula for the length of the cord may be 

l+-^by the binomial theorem 

and integrating the series term by term when a is less 
than each of the quantities l^ and l^' 



Thus, »^=£{\ + '^dx 



-i 



V^2a2 8a^'^T6«6 128a«'^" V 

^ 6a2 40^4 112 «6 1152 a^"^ 

In terms of the deflection, c?, this becomes, since a = -J-, 

4>^ a 



8 



= 1 +?^_?^_L.^^_15^ . 

^ ^ 3/i 5/-J 7 /-J 9 /} 



176 



APPLIED MECHANICS FOR ENGINEERS 



111 like maimer, if c?j = c? + h. 

If the supports are on the same level, l^ = l^ = - and the 

expressions for the total length S become respectively 
1 



a 



- VP + 4 a^ + a^ log, 



Z 4- VZ2 + 4 a2 



"Za 



S=l^l '' 



1 



/s 1 z- 



Z9 



3 23 . a^ 20 25 . «4 56 2' . «« 576 2^ • «» 
TO ; , 23 . 6^2 25 . t/^ , 2 . 2' . r/6 5 . 29 . ^ , 



SI 



5/3 



7/5 



9/' 



For cords for which the deflection is small compared to 
the length of span the series converge rapidly and three 
or four terms are sufficient for computing the length. 
For large values of d the series should not be used. 

Problem 204. A suspension bridge as shown in Fig. 154 has a 
span of 1200 ft. and the cable a niaxiniiuu deflection at the center 
d = 120 ft. The weight of the floor is 2 tons per linear foot. Find 




Fia. 154 



FLEXIBLE COEDS 177 

the equation of the cable and the tension at and at B. If the safe 
strength of cable is 75,000 lb. per sq. in., find the area of wire section 
of cable necessary to support the floor. 

Problem 205. Find the length of the cable in the preceding 
problem. 

Problem 206. A cable is to be suspended from two points distant 
100 ft. ai^art horizontally and 20 ft. vertically. It is to carry a load of 
500 lb. per horizontal foot, and the lowest point of the curve is to be 
25 ft. below the lower point of support. Find the position of the 
lowest point of the cord, the value of the horizontal component of 
the tension, and the maximum tension. 

Problem 207. Find the length of the cable of the preceding 
problem. 

Problem 208. A cable 105 ft. long, carrying a uniform horizontal 
load, is stretched between two points on the same level distant 100 ft. 
apart. Find the deflection. 

Suggestion. Use the first three terms of the series which expresses 
S in terms of / and d. See how large the fourth term is with the 
computed value of d. 

Problem 209. A uniform wire weighing ^ lb. per foot of length is 
supported between two points 200 ft. apart on the same level and the 
maximum deflection is 2i ft. Find the horizontal tension, the 
maxinmm tension, and the length of the wire. 

Suggestion. Since the deflection is here relatively small, the load 
is very nearly uniformly distributed along the horizontal. The curve 
may then be regarded as approximately a parabola. 

100. Equilibrium of a Flexible Cord Carrying a Load Uni- 
formly Distributed along the Cord. — Using a notation like 
that of Art. 97, the same method leads to an equation 
like (5) of that article, with x replaced by s; namely, 

(It/ _ w 

tx ~ X^' 

N 



178 



APPLIED MECHANICS FOR ENGINEERS 



in which the origin is at a point of the curve where the 
slope is zero and s is the length of the curve from the 
origin to the point (a;, ?/). 

Y 

-lo- 

-I ^-1 

As. 








(a) 



(b) 
Fig. 155 



(c) 



X 



Replacing dy by ^ ds^ — dx^ and writing — = a, this 



w 



becomes 



ds^ — dx^ 8^ 



dx^ 



or 



ds 



ds = - Va^ + s^ • dx. 
a 

dx 



Integrating, 



Va2 + s2 « 
loge (s + V s-^ + a^) = _ -^ ^. 



a 



s = when 2: = ; .•.(?= log a, 

log. --^- = - • 

\ a J a 

s H- Va^+ g'^ — ^~. 



Invert, 



a 



« + Va2 + §2 



e ". 



FLEXIBLE CORDS 179 

Rationalizing the denominator, 



8 — Va^ + s^ _ _ g-f^ 



Add to 


— ■ = ea. 

a 


Then 





(1) 



It was found above that -^ = -. 

dx a 



dy = I (e" — ^ a)c?2; 



a 



X 



from which V =■ ~ (^" + ^ «) H- c'. 



9 



But a: = when ?/ = ; . •. c' = — «. 

.•.2/ + a = ^(e« + e"«)> (2) 

which is therefore the equation of the curve, the origin 
being at the lowest point of the curve (Fig. 155 (a) and 

This curve is called the catenary. 

The length of the curve follows at once from equation 
(1) ; namely, 

OA = s^ = ~(ea — e a) 

and OB=s.^ = ^(iei-e~a), (Fig. 155) 

and S = S2± s^ according as the cord hangs as in (a) or (^) 
of Fig. 155. 



180 APPLIED MECHANICS FOR ENGINEERS 

If the points of support are on the same level, 1^ = 1^ = -, 
and the total length of cord is 

The values of y and s may be expressed in a series by 
making use of the expansion 

/y*A /yHJ ^^ /y*0 

e. = l + . + _ + _ + „ + _+.... 
It is left for the student to show that 

c, — / 4- 1 ^1 1 1 ^1 1 

101. Representation by Means of Hyperbolic Functions.— 
From the definitions 

sinh x = \ (^^ — e"""), 
cosh x = \ (e^ + e~^), 

it follows at once that the values of y and «, the arc from 
to the point (ic, ?/), are given by the formulae 

y -\- a = a cosh-, 
a 

8 = a sinh-, where a = — • 
a w 

A table of hyperbolic sines and cosines is given in the ap- 
pendix and should be used in the solution of the following 
problems. 

Problem 210. A flexible wire weighing \ lb. per foot is supported 
by two posts 200 ft. apart. The horizontal pull on the wire is 500 lb. 
Find the deflection at the center and the length of the wire. 



FLEXIBLE CORDS 181 

Problem 211. What pull will be necessary in Problem 210 so that 
the greatest deflection will not be greater than 6 in.? What is the 
length of the wire for this case? 

Problem 212. Find the tension in the wire of Problem 210 at the 
supports. 

Problem 213. A wire weighing \ lb. per foot is suspended be- 
tween two points A and B where B is 20 ft. higher than A and distant 
120 ft. horizontally. The horizontal component of the tension of the 
wire is 50 lb. Find the position of the lowest point of the curve, the 
deflection, the length of the wire, and the maximum tension. Sketch 
the curve in which the wire hangs. 

Suggestion. Assuming that the wire hangs as in (a), Fig. 155, 
we may write 

d + 20 = 200 cosh ^, since a = - = 200, 
200 w 

rf = 200 cosh i. 



Subtracting, 20 = 200 (cosh -|- - cosh ^\ 



= 400sinh'-^sinh'-aj=^', 

since cosh x — cosh y = 2 sinh — ^^—^ sinh ~ -^ • 

2 2 

But ^ + /a = l-'^- 

Therefore 20 = 400 sinh .3 • sinh ^-^:^, 

400 

from which the value of l^ — l^ may be found from the table of hyper- 
bolic sines. Combining this value with ^2 + ^1= 120, the values of /, 
and /g are determined. The values of d, S, and the maximum tension 
may then be determined. 

Problem 214. Solve the preceding problem if B is 40 ft. higher 
than A<, all other conditions remaining unchanged. 



182 APPLIED MECHANICS FOR ENGINEERS 

Problem 215. A wire is suspended between two points on the 
same level 240 ft. apart. The deflection at the center is 60 ft. Find 

the value of «, or — , on the supposition that the load is (1) uniformly 

IV 

distributed along the horizontal, (2) uniformly distributed along the 
wire. Compute the values of // for x = iO and for a; = 80 ft. for both 
cases and sketch the parabola and the catenary. 

Suggestion. In case (1) a is obtained by substituting the coordi- 
nates of the point of support in the equation of the parabola. In case 
(2) the value of a found for the parabola may be used as a trial value 
and the correct value of a found by modifying the trial value until a 
value is found to satisfy the equation 

y -\- a = a cosh - , where // = 60 and x = 120. 
a 



That is, a must satisfy 



h 1 = cosh -^^ 

a a 



Problem 216, Solve the preceding problem if the deflection in the 
middle is only 6 ft. 



CHAPTER IX 

MOTION IN A STRAIGHT LINE (RECTILINEAR MOTION) 

102. Velocity. — When a particle moves along a straight 
line passing over equal spaces in equal times, it is said to 
have uniform motion. In this case the ratio of the space 
passed over to the time taken to pass over that space is 
called the velocity of the particle. 

If the motion is along a straight line but is not uniform, 
the ratio of the S2:)ace passed over in any time to the time 
is called the average velocit}" of the particle for that time, 
or space. Thus, if the particle moves from P to Pj, a 

As 

distance As, in the time A^, then — = average velocity of 

the particle between P and Py 

As 

The limiting value of — as A^ approaches the limit zero 

is defined to be the velocity of the particle at P. 
Thus, .= §5 

tliat is, tJie velocity is the first derivative of the distance tvith 
respect to time. 

The unit of velocity is the unit of space per unit of time, 
as /it. per sec, mi. per hr., etc. Speed is sometimes used 
instead of velocity, especially in speaking of the motion of 
machines or parts of machines. Speed, however, involves 
only the rate of motion without reference to its direction, 
while velocity involves both rate of motion and the direc- 
tion in which the motion takes place. 

183 



184 APPLIED MECHANICS FOR ENGINEERH 

Problem 217. Tlie space passed over by a particle moving in a 
straight line is given by the formula .s = 16 /-, where s is the distance 
in feet passed over in the time t in seconds. Find the velocity wlien 
t = Q, when / = 2, and at the end of / sec. What is the velocity of the 
particle when it has moved 40 ft. V 

Problem 218. A particle moves back and forth along a straight 
line, the abscissa of the particle being given by x — k cos (ct). Find 

the location of the particle and its velocity when / = 0, — , -, '-^ . 

'2 c c c 

pA-ect ordinates to represent the velocity for several positions of the 

moving particle and represent the velocity by a curve. 

103. Acceleration for Straight-line Motion. — When the 
velocity of a particle moving in a straight line increases 
by equal amounts in equal times, the motion is said to be 
uniformly/ accelerated^ and the gain in velocity per unit of 
time is called the acceleration of the particle. 

If, for straight-line motion, the velocity changes from v 

at the time t to v -\- Av at the time t -f- A^, then — is called 
the average acceleration of the particle for the time A^.' 

/\ 9) (111) 

The limitinof value of — , i.e. — , is defined to be the 

At dt 

acceleration of the particle at the time t. 

Thus, if a represents acceleration, 

~ dt~ dP 

Another form for a is frequently used. Since a = — 

and V = — , there results 
dt 

vdv = ads 

by eliminating dt. 



MOTION IN A STRAIGHT LINE 185 

The unit of acceleration is the unit of velocity per unit 
of time, as ft. per sec. per sec.^ mi. p>er hr. per hr.^ yd. 
per min. per sec, etc. 

Problem 219. Find the acceleration of the particle of Problem 
217 at tlu' times specified. Derive an equation which expresses the 
velocity in terms of the space, and from this equation obtain a 
formula for the acceleration in terms of the space. 

Problem 220. Using t for abscissa and s, v, a, respectively, as or- 
dinates, plot the curves which represent the space, velocity, and ac- 
celeration respectively in terms of the time in Problem 217. Show how 
the ordinate to the velocity curve is related to the slope, or gradient, of 
the space curve at any value of t. Likewise for the acceleration and 
velocity curves. 

Problem 221. Find the acceleration of the moving particle in 
Problem 218. Show that the acceleration is proportional to x. Plot 
curves showing x, v, and a in terms of t. Also a curve showing a in 
terms of x. 

Problem 222. Show that if a curve is plotted, using values of 
time as abscissas and values of velocity as ordinates, the area under 
the curve between two values of the time is equal to the space passed 
over by the particle in that interval of time. 

104. Constant Acceleration. — When the acceleration is 
constant, we have the relation dv = a^c?^, a,, representing 
the constant value of a, and thtrefore 



Jdv = Uc I dt., 



and since 



V = act + VQ'y 

ds 

v = — , 
dt 



I ds = a^ j tdt -{■ VqI dt. 



or 8 = ]a^Vi + Vfjt. 



186 APPLIED MECHANICS FOR ENGINEERS 

In ti similar way the relation 

vdv = a^ds 
gives 

/vdv = aA "ds ; 

V 1'^ 

therefore -y = ^c^, 

or s = 



2ac 

These equations of motion give the velocity in terms of 
time, the distance in terms of time, and the distance in 
terms of velocity. 

105. Freely Falling^ Bodies. — Bodies falling toward the 
earth near its surface have a constant acceleration. It is 
usually represented by ^ and equals approximately 32.2 ft. 
per second per second. The value of g varies slightly with 
the height above the sea level and the latitude, but for 
the purposes of engineering it may usually be taken as 
32.2. The equations of motion for such bodies are, then, 

v = gt-\- Vq, 





s= J^^^ + V' 




^_v^-vl^ 
2^ 


If the body falls 


from rest, Vq =0, and the equations of 


motion become 






v = ot, 








v^ 
^=2,- 



This latter is often written v^ = '2gh, where h = 8. 



MOTION IN A STRAIGHT LINE 187 

106. Body Projected Vertically Upward. — When a body 
is projected verticall}^ upward from the earth, the accelera- 
tion is constant and equals — g. If the velocity of pro- 
jection is Vq, the equations of motion are 

V = —gt + v^t, 

Problem 223. A body is projected vertically downward with an 
initial velocity of 30 ft. per second from a height of 100 ft. Find 
the time of descent and the velocity with which it strikes the ground. 

Problem 224. 'A body falls from rest and reaches the ground in 
6 sec. From what height does it fall, and with what velocity does 
it strike the ground ? 

Problem 225. A body is projected vertically upward and rises to 
the height of 200 ft. Find the velocity of projection y-^, and the time 
of ascent. Also find the time of descent and the velocity with which 
the body strikes the ground. 

Problem 226. A stone is dropped into a well, and after 2 sec. 
the sound of the splash is heard. Find the distance to the surface of 
the water, the velocity of sound being 1127 ft. per second. 

Problem 227. A man descending in an elevator whose velocity 
is 10 ft. per second drops a ball from a height above the elevator 
floor of 6 ft. How far will the elevator descend before the ball 
strikes the floor of the elevator? 

Problem 228. In the preceding problem, suppose the elevator 
going up with the same velocity, find the distance the elevator goes 
before the ball strikes the floor of the elevator. 

107. Newton's Laws of Motion. — Three fundamental laws 
may be laid down whicli embody all tlie principles in 
accordance with which motion takes place. These are 



188 APPLIED MECHANICS FOR ENGINEERS 

the result of observation and experiment and are known 
as Neivtons Laws of Motion. 

First Law. Every body remains in a state of rest or 
of uniform motion in a straight line unless acted upon by 
some unbalanced force. 

Second Law. When a body is acted upon by an unbal- 
anced force, motion takes place along the line of action of 
the force, and the acceleration is proportional to the force 
applied and inversely proportional to the mass of the body. 

Third Law. To every action of a force there is always 
an equal and opposite reaction. 

The second law states that in case the system of forces 
acting on the body is unbalanced, the motion is accelerated. 
Motion takes place in the direction of the resultant force 
with an acceleration proportional to the force. It also 
implies that each force of the system produces or tends 
to produce an acceleration in its own direction propor- 
tional to the force. That is to say, each force produces 
its own effect, regardless of the action of the other forces. 

As a result of this latter fact, if a body is acted upon 
by a force P and the earth's attraction (r^, we have 

P: a=^a:g, 

where G is the weight of the body, ^ the acceleration of grav- 
it}^ and a the acceleration due to the force P. From this it 
follows that Y> ^ ^ Tir ^ 



ri 

The quantity — , in which G- is the weight of the body 

in pounds, and g is in /f. /sec. 2, is called the mass of the 
body in " Engineer's Units.'' 



MOTION IN A STRAIGHT LINE 



189 




Fig. 15(i 



If ^=32.2, the Engineer's I'nit of mass is equivalent 
to 32.2 pounds. 

108. Motion on an Inclined Plane. — A body (See 
Fig. 156), of weight (r, moves down an inclined plane, 
without friction, under the action of 
a force G- sin 6. The acceleration down 
tlie plane equals the accelerating 
force divided by the mass (Art. 107) 

= — -j^ — =g sin 6. The acceleration is 

y 

constant. The equations of motion for such a case, then, 

are (Art. 104) 

1? = (</ sin 9) ^ + Vo, 

2 i/ sine* 

If the body starts from rest down the plane, Vq = 0. If 
it be projected up the plane with an initial velocity Vq, the 
acceleration equals — g sin 6. 

Problem 229. A body is projected up an incHned plane which 
makes an angle of 60° with the horizontal with an initial velocity of 12 
ft. per second. Neglecting friction of the plane, how far up the plane 
will the body go ? Find the time of going up and of coming down. 

Problem 230. A body, weight 20 lb., is projected down the plane 
given in the preceding problem with a velocity of 20 ft. per second. 
How far will it go daring the third second? 

Problem 231. Suppose the body in the preceding problem 
meets a constant force of friction F = 10 lb. What will be the 
acceleration down the plane? How far will it go during the second 
second ? 



190 



APPLIED MECHANICS FOR ENGINEERS 



Problem 232. A boy who has coasted down hill on a sled has a 
velocity oi' 10 mi. per hour when he reaches the foot of the hill. 
He now goes on a horizontal, meeting a constant resistance of 25 lb. 
If the combined weight of the boy and sled is 7.") lb., how far will he 
go before coming to rest? 

Problem 233. Suppose that in the preceding problem the boy 
weighs G5 lb. and the sled 10 lb., and that the boy can exert a force 

of 20 lb. horizontally to keep him on 
the sled. Will the boy remain on 
the sled when the latter stops, or 
will he be thrown forward? 



T 
A 

? Problem 234. Two weights, 

<^'.0 D^'i 1 (7^^51b. andG'2 = 101b. (Fig. 157), 

T attached to an inextensible cord 
G^ which runs over a pulley, are acted 
upon by gravity ; no friction ; mo- 
tion takes place. Find the tension 
in the cord, and the acceleration. 
Consider G^ and G-^ separately with the forces acting upon them, and 
call the tension in the cord T. 



Q. 




Fig. 157 



Then apply the principle " acceler- 
ating force equals mass times 
acceleration.'^ 

Problem 235. A body whose 
weight G = 5 lb. is being drawn 
up an inclined plane as show in 
Fig. 158 by the action of the 
weight G\ = 20 lb. Suppose the 
resistance, F, offered by the plane 

is 10 lb., and that G starts from rest. How far up the plane will G 
go in 6 sec. ? 

Problem 236. An elevator (Fig. 159), whose weight is 2000 lb., is 
descending with a velocity, at one instant, of 2 ft. per second, 
and one second later it has a velocity of 18.1 ft. per second. Find the 




Fig. 158 



MOTION TX A STRAIGHT LINE 



191 




Fig. 159 



tension T in the cable that supports the elevator, 
assuming uniform retardation. 

Problem 237. Suppose the elevator in preced- 
ing problem going up with the same acceleration. 
Find the tension in the cable. 

Problem 238. A man can just lift 200 lb. 
when standing on the ground. How much could 
he lift when in' the moving elevator of the preced- 
ing problems, (a) when the elevator was ascend- 
ing? (h) when descending? 

Problem 239. Two weights, G and G' , are 
connected by an inextensible flexible cord that passes over a friction- 
less pulley, as shown in Fig. 160. G = 20 lb., G' = 100 lb., and there 

is no friction on the plane. 
Find the tension in the cord 
and the acceleration of the 
two bodies. 

Problem 240. A 30- 
ton car is moving with a 
velocity of 30 mi. per hour 
on a level track. The 
brakes refused to work. 
How far will the car go after the power is turned off before coming 
to rest, if the friction is .01 of the weight of the car ? 




Fig. 160 



109. Variable Acceleration. — It has already been shown 

that v = — ^ , a = -~ = — -, and vdv = ads. These relations 
dt dt dt^ 

hold true no matter whether the acceleration is constant 

or variable. If the acceleration is constant, the equations 

of motion are those that have already been worked out 

(Art. 104), and by simple substitution in these equations 

it is possible to find the velocity in terms of the time, the 



192 APPLIED MECHANICS FOR ENGINEERS 

distance in terms of time, and the distance in terms of 
velocity. 

kJ 

If the acceleration is variable, it is necessary to work 
out the equations of motion for each case. This may be 
done, when it is known how a varies, by means of either 
of the equations, 

CL = ^, 

or vdv = ads. 

The latter equation will usually give the beginner less 
difficulty. 

110. Harmonic Motion. — Let it be supposed that a body 
is moved by an attractive force which varies as the dis- 
tance. That is, the attractive force is proportional to the 
distance. Then the acceleration is also proportional to 
the distance. 

Let the acceleration = — ks. 
Then vdv = — ksds, 

vdv = — k \ sds ; 

therefore v^ — vl = — ks\ 

where Vq is the initial velocity when s equals zero and 
k is the factor of proportionality, determinable in any 
special case. This equation gives the relation between 
the velocity and distance. Since v — Vv§ — ks^^ it is 
evident that v = when >Jk ■ s = Vq. This means that the 



body comes to rest when s lias reached a certain value, 

v^ 
viz. ,y' From the original assumption, a — — ks, it is 



MOTION IN A STRAIGHT LINE 193 

seen that the acceleration is greatest when s is greatest, 

that is, when s = —= ; and is least when s is least, that is, 

when 8=0. 

To get the relation between distance and time, the 
equation v = Vv^ — ks^ may be put in the form 

ds 
. == =dt^ 

V ^5 — k^^ 

from which — =- sin~i '— = f, 

or IT sin ^kt = s. 

This relation between the distance and time shows that 

as t increases s changes in value from — p to ~7=^ , as- 

wk \k 

suming all values between these limits, but never exceed- 
ing them, since sin ^kt can never be greater than + 1 or 
less than —1. The motion is, therefore, vibratory or 
periodic, and is known as harmonic motion. The complete 

period in this case is '^^^-^' 

The relation between velocity and time may be found 
for this case by differentiating the last equation with 
respect to time. Then, 

V = Vq cos ^kt 

This shows that v^ is the greatest value of v. 

This motion is usually illustrated by imagining a ball 
attached to two pins by means of two rubber bands or 



194 APPLIED MECHANICS FOR ENGINEERS 

springs, since the force exerted by either of these is pro- 
portional to the elongation. (See Fig. 161.) Assuming 
^ ^ that there is no friction and 

L ^ that tlie ball is displaced to 

±10 a position B by stretching 

^'^' ^^^ one of the rubber bands, 

when released it continues to move backward and forward 
with harmonic motion. 

Problem 241. Suppose the ball in Fig. 161, held by two helical 
springs, to have a weight of 10 lb. and that it is displaced 1 in. 
from 0. The two springs are free from load when the body is at O. 
The springs are just alike, and each requires a force of 10 lb. to 
compress or elongate it 1 in. Find the time of vibration of the 
body and its velocity and position after ^- sec. from the time when 
it is released. 

It has been found by experiment that the force necessary to com- 
press or elongate a helical spring is proportional to the compression 
or elongation, 

111. Motion with Repulsive Force Acting. — Suppose the 
force to be one of repulsion and to vary as the distance ; 
then a = ks^ and vdv — ksds^ so that 



These equations show that as t increases s also increases 
and the body moves farther and farther away from the 
center of force. The motion is not oscillatory. 

112. Motion where Resistance Varies as Distance. — I f a 
body whose weight is 644 lb. falls freely from rest through 
60 ft. and strikes a resisting medium (a shaft where fric- 



MOTION IN A STRAIGHT LINE 



195 



tion on the sides equals 2 ^= 10 times the 
distance ; see Fig. 162), since accelerating 
force equals mass times acceleration, 



a = 



a-2 F 
M 



a-\os s 

9 



It is required to find (a) the distance 
tlie body goes down the shaft before 
coming to rest ; (l>^ the distance at which 
the velocity is a maximum ; (c) the total 
time of fall ; (c?) the velocity at a distance 
of 10 ft. down the shaft. 

After striking the shaft the relation be- 
tween velocity and distance is as follows : 



£vdv^j^(g-fj,s. 




Fig. 162 



The remainder of the problem is left as an exercise for the 
student. 

Problem 242. A ball wliose weight is 32.2 lb. falls freely from 
rest through a distance of 10 ft. and strikes a 400-lb. spring (Fig. 
16;i). Find the compression in the spring. It is to be miderstood 
that a 400-lb. spring is such a spring that 400 lb. resting upon it 
compresses it one inch, and 4800 lb. resting on it compresses it one 
foot, if such compression is possible. After the ball strikes the spring 
it is acted upon by the attraction of the earth and the resistance of 



the spring. The acceleration n is then 



^-4800.^ 
M 



, where s is meas- 



ured in feet. The relation between velocity and distance is then ob- 
tained from the relation, 

Jv„ Jo 



196 



APPLIED MECHANICS FOR ENGINEERS 



Problem 243. A 20-ton freicrht 
car (Fig. 104), moving with a 
velocity of 4 mi. per hour, strikes 
a bumping post. The 6U,000-lb. 
spring of the draft rigging of the 
car is compressed. Find the com- 
pression s. Assume that the 
bumping post absorbs none of 
the shock. 

Problem 244. Suppose the 
car in the preceding problem to 
be moving with a velocity of 4 
mi. per hour, what should be 
the strength of the spring in the 
draft rigging so that the com- 
pression cannot exceed 2 in. ? 

Problem 245. After the 

spring in Problem 242 has been 

compressed so that the ball comes 

to rest, it begins to regain its 

Fig. 163 original form. Find the time 

required to do this and the velocity with which the ball is thrown 

from the spring. 




4 Ml. PER HR. 
< 




izzWaaamaad 




Fig. 164 



MOTION IN A STRAIGHT LINE 



197 



113. Motion when Attractive Force Varies Inversely as the 
Square of the Distance. — This is the 

I case of motion (Fig. 165) vvheu two 

Q bodies in space are considered, since in 

such cases the attractive force varies 
directly as the product of their masses 
and inversely as the square of the dis- 
tance between their centers of gravity. 
The same attraction holds between two 
opposite poles of magnets or between 
two bodies charged oppositely with 
Fig. 165 electricity. 

— k 
Suppose the acceleration = — — and that the initial 




velocity is zero. 



Then, j vdv = — 1 



■'"^ 



C?8, 



SO that 



and 



cjs__ \'lk_ Vg(^.s 

dt ~ ^' Sr 



S2 



'0 



Sr 



^2k 



■VsqS 



52 — Q vers 



-1 






The time required to reach the center of attraction 
from the position of rest is obtained by putting s = 0. 



This gives t = — I -^ 

It is seen that when s = 0, the velocity is infinite, and 
therefore the body approaches the center of attraction 
with increasing velocity and passes through the center, to 
be retarded on the other side until it reaches a distance 



— 8q. The motion will be oscillatory. 



198 APPLIED MECHANICS FOR ENGINEERS 

If one of tlie bodies is the earth, of ladius r, and the 
other is a body of weight G- falling toward it, the equa- 
tions just derived hold true. In this case it is possible 
to determine k. The attraction on the body at the sur- 
face of the earth is 6r, and at a distance s is F^ so that 

F = Grl —]. The acceleration is therefore -— — = — g{ — 
\s^J M ^ W. 

This gives k^ then, equal to r^g. 

Substituting these values in the above equation, we find 



v = yJEE 



2 VSqS — s^ 



When s = r at the earth's surface. 



V = yj'^ V(So - r) = V2 gr^fl 



r 



Sq \ 8( 



If 8q= cc^ v = V2 gr. 

But this is a value of v that cannot be obtained, since 
8q cannot be infinite. So that the velocity is always less 

than V2^r. It is interesting to notice here that if a 
body were projected from the earth with a velocity greater 

than V2^r, it would never return, provided there were 
no atmospheric resistance. Substituting ^ = 32.2 and 
r = 3963 mi., 

V = 6.95 mi. per sec. 

This is the greatest velocity that a body could possibly 
acquire in falling to the eartli, and ii body projected 
upward with a greater velocity would never return 
(neglecting resistance). 



MOTION IN A STRAIGHT LINE 199 

If the body falls to the earth from a height A, the veloc- 
ity acquired may be obtained from the foregoing by put- 
ting s = r and s^^ = li -\- r \ then 



I'lgrh 



If h is small compared to r, this may be written, without 
serious error, 

which is the formula derived for a freely falling body in 
Art. 105. 

Problem 246. A body of mass 10 lb. has an initial velocity of 50 
f /s and moves in a medium which resists with a force proportional to 
the velocity and equal to 1 lb. when the velocity is 4 f /s. How far 
will the body move before the velocity is reduced to 10 f /s? Would 
the velocity ever become zero with that law of resistance ? What 
would be the limiting value of the space passed over? 

Problem 247. A body of weight W lb. is projected vertically 
upward with a velocity I^o- If the resistance of the air is equal to kv'^, 
prove that the height to which the body will rise is 

,.JElog.(l+^o). 
2 kg ^ V W/ 

Problem 248. The body of the preceding problem falls again to 
the earth. Show that the velocity with which it reaches the earth is 



Problem 249. A man jumps from a balloon and acquires a 
velocity of 80 f /s before his parachute is fully opened. The man and 
parachute weigh 150 lb. The resistance of the air varies as the 
square of the velocity (approximately) and is 1 lb. per square foot of 
opposing surface \vhen the velocity is 20 f /s. If the diameter of the 
parachute is 12 ft., what velocity will tlie man have after descending 
a further distance of 200 ft. ? after 1000 ft. ? 



200 



APPLIED MECHANIC ti FOli ENGINEERS 



Problem 250. On a toboggan slide of constant slope assume the 
friction to be constant and the resistance of the air to be proportional 
to the square of the velocity. Derive the formula for the velocity in 
terms of the distance, starting from rest. 

114. Relative Velocity. — When we speak of the velocity 
of a body, it is understood that we mean the velocity of 
the body relative to the earth, more particularly the point 
on the earth from which the motion is observed. Since 
the earth is in motion, it is evident that velocity as gen- 
erally spoken of is not absolute velocity, and since there 
is nothing in the universe that is at rest, all velocities 
must be relative. In everyday life, however, we think of 
velocities referred to any point on the surface of the earth 
as being absolute. 

Suppose two particles, A and B^ to have the velocities 
Va and Fft, as illustrated in Fig. 166. 77ie motion of B 
relative to A is obtained by regarding A 
as at rest and B as having a velocity com- 
posed of the actual velocity of B and the 
reversed velocity of A at each instant. 
Thus, the vector F", which is the result- 
ant of Vb ^i^d ^ reversed, is the velocity 
of B relative to A at the instant at which 
the velocities are Va ''^^^^ ^"6- 

This is made more evident in the case 
of constant velocities by the following 
consideration : if F^ and Vj, are constant, then at the end 
of a unit of time A would be in the position A^^ and B in 
B^, and their distance apart and direction from each other 
would be represented by the line Aj^B^^ while if A re- 




FiG. 16G 



MOTION IN A STRAIGHT LINE 



201 



mained at rest, and B moved with a velocity composed of 
Vj, and Va reversed, the line AB' would represent the dis- 
tance and direction of B from A. Since A^B^ and AB' are 
evidently equal and parallel, the relative positions of the 
two particles w^ould be the same in one case as in the other. 

As an illustration, consider the velocity of the wind 
relative to the sail of an ice boat. Let F^ and V^^ be the 
velocities of the 
boat and wind re- 
spectively (Fig. 
167) and SL the 
direction of the 
sail, wd:iich is as- 
sumed to be a 
smooth plane sur- 
face. 

Combining F^ 
with Vb reversed, 
the velocity V is 
obtained, which is 
the velocity of the 
wind relative to the boat. This velocity can be resolved 
into two components, OA and 0^, along and perpendicular 
to the sail respectively. The component OA will have no 
effect on the boat on the assumption that the sail is a 
smooth surface. The component OU may be resolved into 
two components, OB and 00, respectively, along and per- 
pendicular to the path of the boat. The vector OB then 
represents the component of the velocity of the wind which 
urges the boat forward. 




Fig. 167 



202 APPLIED MECHANICS FOR ENGINEERS 

From the figure it is clear that there may be a forward 
component of the wind's velocity on the boat even when 
the velocity of the boat is greater than the velocity of the 
wind. It is only necessary for the velocity of the wind 
relative to the boat to fall in front of the sail. Then as 
long as the forward pressure of the wind is greater than 
the resistance of the ice the velocity of the boat will 
increase. In the case of ice boats, where the resistance 
is small, the velocity of the boat may greatly exceed the 
velocity of the wind for high wind velocities. 

Problem 251. A train is moving with a speed of 60 mi. per 
hour, and another train on a parallel track is going in the opposite 
direction with a speed of 40 mi. per hour. What is the velocity 
of the second train as observed by a passenger on the first? 

Problem 252. A man in an automobile going at a speed of 40 
mi. per hour is struck by a stone thrown by a boy. The stone has 
a velocity of 30 ft. per second and moves in a direction perpendicular 
to the direction of motion of the automobile. With what velocity 
does the stone strike the man? 

Problem 253. A man attempts to swim across a river, \ mi. 
wide, which is flowing at the rate of 3 mi. per hour. If he can swim 
at the rate of 4 mi. per hour, what direction must he take in swim- 
ming in order to reach a point directly across on the opposite shore. 
What distance will he swim relative to the water ? How long will it 
take for him to cross ? 

Problem 254. An ice boat is moving due north at a speed of 60 
mi. per hour, and the wind blows from the southwest with a velocity 
of 40 mi. per hour. What is the apparent velocity and direction of the 
wind as observed by a man on the boat? 

Problem 255. A man walks in the rain with a velocity of 4 
mi. per hour. The raindrops have a velocity of 20 ft. per second in a 
direction making 60° with the liorizontal. How much must the man 
incline his umbrella from the vertical in order to keep off the rain : («) 



MOTION IN A STRAIGHT LINE 203 

when going against the rain, (b) when going away from the rain ? If 
he doubles his speed, what change is necessary in the inclination of his 
umbrella in (a) and (b) ? 

Problem 256. The light from a star enters a telescope inclined 
at an angle of 45° with the surface of the earth. The velocity of 
light is 186,000 mi. per second and the earth (radius 4000 mi.) makes 
one revolution in 24 hr. What is the actual direction of the star with 
respect to the earth ? This displacement of light due to the velocity 
of the earth and the velocity of light is known as aberration of light. 

Problem 257. A locomotive is moving with a velocity of 40 mi. per 
hour. Its drive wheels are 80 in. in diameter. What is the tangential 
velocity of the upper point of the wheels with respect to the frame of 
the locomotive? What is the tangential velocity of the lowest point? 

Problem 258. Show that if an ice boat were moving north and 
the wind had a velocity of 20 mi. per hour from the southwest, the 
maximum velocity that the boat could attain if there were no friction 
would be 38.6 mi. per hour if the sail w'ere set at 30° with the direction 
of the boat's motion. What effect would increasing the angle between 
the sail and the boat's direction have in this case? 

Problem 259. With the sail set at 30° with the direction of the 
boat's motion, in what direction could the boat go fastest if there 
were no friction? Ans. AVhen the direction of the boat makes an 
angle of 60° with the direction of the wind. 

Problem 260. Given that the angle which the sail makes with 
tlie direction of motion of the ice boat is SO"^, the angle which the 
direction of the wind makes with the direction of the boat's motion 
is 45°, the area of the sail is 50 sq. ft., the resistance of the ice 
to the boat's motion is 16 lb., and given that the pressure of the 
wind normal to the surface of the sail varies as the square of the 
velocity (component of relative velocity normal to sail), and is 1 lb. 
per square foot of opposing surface when the velocity is 15 mi. per 
hour, find the maximum velocity that the boat can attain when the 
velocity of the wind is (a) 20 mi. per hour, (b) 30 mi. per hour. 

Ans. (a) 14.6 mi. per hour. (/;) 31.0 mi. per hour. 



CHAPTER X 



CURVILINEAR MOTION 



115. Velocity. — Suppose a 
particle to be moving along 
the curve of Fig. 168, from 
A toward B. The average 
velocity of the particle be- 
tween A and B is defined as 
the velocity the particle 
would have if it moved uni- 
formly from A to B, i.e. 
along the chord AB with con- 
stant speed. The average 
velocity of the particle between A and B is therefore, 

1 ., chord AB 
average velocity = -— - • 

time spent between A and B 

If the chord is represented by Ac and the time by At, 
then 




Fig. 168 



average velocity = — , 



Ac 



and is represented by a vector A (7, of length — , laid off 

from A on AB. When the time At is made to approach 

Ac 
the limit zero, the limiting" value of the ratio — is defined 

^ At 

to be the velocity of the particle at the point A. 

204 



CURVILINEAR MOTION 205 

The limiting direction of the chord is the tangent at A. 
Hence, calling v the velocity of the particle at the point A^ 

dc 

V = — 

dt 

dc 

and is represented by a vector of length — laid off from 

A along the tangent in the direction of motion. 

It is shown in calculus that if a single tangent to the 
curve at A exists, then the limit of the ratio of the arc to 
its chord, as the arc approaches the limit zero, is unity. 
Hence, if tlie arc AB is As, then 

c?s _ 1 

dc 

1 dc ds ds 

and v = — — = —-' 

at dc dt 

The velocity of a particle moving along a curved path is 
therefore represented at any point of the path by a vector 

equal to the value of — laid off on the tangent to the path 

dt 

at the given point, and in the direction of the motion. 

116. Acceleration in a Curved Path. — Let the velocities 
at A and B (Fig. 169 (a)) be v and v -\- Av respectively. 
Lay off the vectors representing these velocities from the 
same point A (Fig. 169 (^)). 

The vector CD, from the end of v to the end of v -\- Av, 
represents the change in velocity in the time A^. Tlie 
ratio of this vector to the time At is the average accelera- 
tion for the time A^. Thus the vector CJE represents the 
average acceleration of the particle between A and B. 



206 



APPLIED MECHANICS FOR ENGINEERS 



The limiting value of this average acceleration as A^ 
approaches zero as a limit is defined to be the acceleration 
of the particle at A. Letting v, and v, be the projections 




Fig. 169 



of V upon rectangular axes, then the projections of CD 
on these axes are Av^ and Avy and the projections of CU 

are ^^ and ^^ Hence 

A^ A^ 



and calling a the acceleration of the particle at A, 






\dtj \dtj 
If a and a^ are the components of a along the axes, then 



dv. 



dv. 



dt 



a,. = 



iiiw. 



dt 



Now 



v^ = v cos and Vy = v sin 6 



CURVILINEAR MOTION 



207 



where is tlie angle which the tangent to the curve at A 
makes with the .i^-axis. Therefore 



a^ = cos D V sm u — , 

dt dt 



• ndv , ^ 

a^ = sm u —- + V cos 6 



dt 



d6 
dt 



Since 



V = 



ds 
dt 



dt 



dt^ 



dO 



A value of ^ may be found as follows : Draw a normal 

dt 

to the curve at A (Fig. 170). On this normal there is a 
point 0', the center of a circle tangent to the curve at A 
and passing through B. Let r' be the radius of this circle 
and A^' the angle 



subtended at 0' by 


0' 




the arc AB. As B 


i^^Nv 


j / 


is made to approach 
^ as a limit the tan- 


\\aAO \s^ 


¥ 


gents to the curve 
and the circle at B 


\ 


"^^/a^,^^^ 


take the same limit- 


\\ ^ 


\ 


ing position and 


\\^^y 


: 


Fig. 


170 



Ai9' 
approaches 1 as a limit. The ratio of the arcs, As' of the 
circle, and As of the curve, has the limit 1, since they have 
the same chord and the limit of the ratio of each arc to its 
cliord is 1. 

As the point B is made to coincide with A the point 0' 
takes some limiting position, 0, wliich is called the center 
of curvature of the curve at A. If r is the value of Ovl, 
then limit r' = r. 



208 APPLIED MECHANICS FOR ENGINEERS 

Now r'AO'^As', or^ = l. 

As' r' 

rrtu £ dO' 1 . d6 ^ 1 6?S -, 

Ineretore — — = -, or since — - = 1 and — - = 1, 

ds r du' as' 

^ = 1 ^dl^dSd^^lch^v^ 
ds r dt ds dt r dt r 

The quantity r is the radius of curvature of the curve at A. 

Substituting the values found for — and ^, the values 
01 a^ and ay become 

ax = cos e ^— 9 

^ - oi,. A ^^"* ^ ^^ cos 9 
<*« = sin y — — - H • 

117. Tangential and Normal Components of the Accelera- 
tion. — The above values of a^ and ay hold for any rectangu- 
lar axes. Let the a^-axis be taken along the tangent at A 
and the «/-axis along tlie normal. Then cos^=l, and 
sin ^ = at the given point, and the values of a^ and ay be- 
come respectively 

The normal force and tangential force may now be 

written, 

Normal force = ^^^ ; 
r 

Tangential force = 3r ^. 

For all curves except the circle r, the radius of curvature 
varies from point to point. In the circle, however, it is the 
radius, and is therefore constant. Tn this case the normal 
force is usually called tlie centripetal force. 



CURVILINEAR MOTION 



209 



Problem 261. A ball weighing 2 lb. is attached to one end of a 
string 5 ft. long, the other end of which is attached to a fixed point. 
The ball is pulled aside and released. When the string makes an angle 
of 30'^ with the vertical, the velocity of the ball is 10 f/s. Find the 
normal and tangential components of the acceleration and the accel- 
eration in magnitude and direction at that instant. Find also the ten- 
sion in the string. (Notice that the force acting tow^ard the center on 
the body is the difference between the tension in the string and the 
component of the weight along the direction of the string.) 



118. Uniform Motion in a Circle. — A body moving with 
constant speed, v^ in the circumference of a circle is acted 
upon by one force, the normal or centripetal force, and this 



equals 



That this is true is evident when it is re- 



membered that the tangential velocity is constant, thus 
making the tangential accelera- 
tion zero. An illustration of 
uniform motion in a circle is 
seen in the case of the simple 
governor shown in Fig. 171. 
When the speed is constant, 
then a, A, and r are constant. 
Let T be the tension in the rod 
supporting the ball, then, since 
there is no vertical motion 
2 y = 0, so that T cos « = a. 




Fig. 171 



Jii;2 



Considering the normal force, we have T sin a = , so 

r 

thattana=— . From these equations 7^ may be found 
for any values of a and r. 



210 



APPLIED MECHANICS FOE ENGINEERS 



Problem 262. The 
weighted governor shown 
ill Fig. 172 is rotated at 
such a speed that a =30°. 
Find the forces acting on 
the longer rods and the 
stress in the shorter rods. 
The connections are all 
pin connections. 

Problem 263. A type 
of swing is shown in Fig. 
17o. A revolving central 
post supported by wires ^l 
and B carries six cars G, 
each suspended from cross 
arms D by means of cables 
50 ft. long. When the 
swing is at rest, the cars 
hang vertically and a = 0; 
as the speed of rotation increases, a becomes larger. Suppose the car 
and its load of four passengers to weigh 1000 lb., and the speed to be 
such that «;= 30°; find the tension in the cables supporting the cars. 
Assume that a single car is carried by one cable. 

Problem 264. The 

same principle that 
has been seen to hold 
for motion in a circle 
enables us to solve a 
problem that comes 
up in railroad work. 
When a train goes 
around a curve, it is 
desirable to have the 
outer rail raised suffi- 
ciently so that the 




Fig. 172 




CURVILINEAR MOTION 



211 



wheel pressure will be normal to the rails. It is really the same 
problem as Problem 263, where the sustaining cable is replaced by 
a track. (See Fig. 174.) Let r be the radius of curvature, v the veloc- 
ity of the car, of weight G. Show that the superelevation of the 

outer rail is given by tan a = — , and 

so, approximately, 

h= — , 

gr 

where cl is the distance between the 
rails in feet, r the velocity in feet per 
second, g is 32.2, ;• is the radius of 
curvature in feet, and h is the superelevation of the outer rail in feet. 

Problem 265. Show that, using d = 4.9 ft., this height may be 
expressed, approximately, as follows : 




Fig. 174 



h = 



3r 

where h and r are in feet and i\ is the velocity in miles per hour. 

Problem 266. Find the superelevation of the outer rail on a 
curve of radius 2000 ft. for speeds of 20 mi. per hour, 40 mi. per hour, 
60 mi. per hour. 

Problem 267. What would the superelevation need to be for the 
above speeds on a curve of radius 500 ft. ? 

119. Motion without Friction along" Any Curve in a Vertical 

Plane. — Let a 
particle slide 
down the smooth 
curved track 
(Fig. 175), start- 
ing from rest at 
the point A dis- 
FiQ. 175 tant 1/q above a 




212 APPLIED MECHANICS FOR ENGINEERS 

horizontal line of reference. When in the position P, the 
forces acting on the particle are the weight (r, acting ver- 
tically downward, and the force exerted by the track, 
which is normal to the track since there is no friction. 
The tangential force is then Gr sin 6, and hence for de- 
scending motion, 

---= a sin (9, 
g dt^ 

where s is the length of the curve AP measured from A 

toward P. 

XT • /} dy -y -i (Ps dv 

Now sm u = — f- and also = v --. 

ds dt" ds 

mi £ dv dy 

Ihereiore v— - = — g^. 

ds ds 

Integrating, -=: -gg-\- C. 

When y =iy^^ v=0 \ therefore C = gy^. 

Therefore 

Hence, the velocity acquired by a body in sliding down a 
smooth track is the same at any point as if the body had 
fallen freely through the same vertical distance. 

The time of descent depends upon the form of the curve. 
Problem 268. Show that the equation 

vdv — — gdy 

holds when the body is ascending as well as when descending, and 
then prove that the body would rise on the track to the level from 
which it started. 



CURVILINEAR MOTION 



213 



Problem 269, A body of weight G starts from rest at the top of 
a smooth circular track of radius R in a vertical plane. Find the 
velocity, acceleration, and force exerted by the track on the body 
when it has traveled through 60°, 90°, 150°, 180°, 210° of arc. 

Problem 270. AVith what velocity would a body have to be pro- 
jected from the lowest point of a smooth circular vertical track in 
order that its velocity would just keep it from falling from the track 
at the top? 

Problem 271. If the body of the preceding problem were pre- 
vented from leaving the track, what velocity at the lowest point 
would just carry it around the track ? 

Problem 272. In the centrifugal railway (Fig. 176), friction 
neglected, what would have to be the ratio of h to A' so that the car 
would not leave the track at .4, starting from rest at the height h ? 




Fig. 17(i 



Problem 273. If the car in the preceding problem were prevented 
from leaving the track, what would be the ratio of h to h' to just 
carry the car past A ? 

Problem 274. A particle slides from rest from any point on a 
smooth sphere. Show that it will leave the sphere when it has de- 
scended vertically through one third of its original vertical distance 
above the center. 



214 



APPLIED MECHANICS FOR ENGINEERS 



Problem 275. A locomotive weighing 175 tons moves in an 800- 
ft. curve with a velocity of 40 mi. per hour. Find the horizontal pres- 
sure on the rails, if they are on the same horizontal. 

Problem 276. If the velocity of the earth was 18 times what it 
actually is, show that the force of gravity would not be sufficient to 
keep bodies on the earth near the equator. Take the radius of the 
earth as 4000 mi., and assuming the above conditions, find at what 
latitude the body would just remain on the earth. 

Problem 277. A pail containing 5 lb. of water is caused to swing 
in a vertical circle at the end of a string 3 ft. long. Find the velocity 
of the pail at the highest point so that the water will remain in the 
pail. Find also the velocity of the pail at the lowest point. 

120. Simple Circular Pendulum. — The simple circular 
pendulum consists of a weight Gr suspended by a string 

without weight, of 
length Z, in such a way 
that it is free to move 
in a circle in a vertical 
plane due to the action 
of gravity. (See Fig. 
177.) Let B be such a 
position of the pendu- 
B lum that its height 
above the horizontal is 
h, and (7 any other po- 
sition designated by the 
coordinates x and 7/. 
Let the weight be G 
and the tension in the string T. These are the only forces 
acting on the body. The only forces that can produce 
motion in the circle are those that are tangent to the circle 




G 



Fk;. 177 



CURVILINEAR MOTION 215 

OB. The force T is normal to the circle and so has no 
tangential component. The force G has a tangential com- 
ponent — Gr sin 0. The equation of motion is, therefore, 

vdv = ads = — g sin Ods, 

arc (J \ 9 
where ^ = — =-, where s denotes distance along 

the curve. 

Since s = 16, ds — Idd, 
and hence vdv = — Ig sin 6d6. (1) 

Approximate Solution. The integration of this equa- 
tion leads to a complicated relation between the angle 
and the time. An approximate solution can be obtained 
for small angles of oscillation by replacing sin 6 by 6. 
Equation (1) then becomes 

vdv = - IgOdd. 
Integrating, v^ _ _ /^^2 _|_ q^ 

Let v= when 0=9^; then C=^lge\. 



.-. v= ^Ig-VOl-d^. 

For the body descending s decreases as t increases and 

therefore — is neo^ative, and 
dt 

dt dt ^1 



dO 
or 



V6>2 - e^ 



^-yR-dt. 



Integrating, sin~^— = — \-t+ Cj. 



216 APPLIED MECHANICS FOR ENGINEERS 

Let ^ = when 6=0^; then sin~^l = C^ 
From the angles whose sines are 1 choose — = sin~U= Oy 

Then ^.« = |_sin->^. (2) 

As the pendulum swings from the highest point on the 

right to the point on the left on the same level (Art. 119), 

i.e. to where = — 6^, the value of t continually increases. 

9 
Equation (2) shows that sin"^— must continually decrease 

}_ 
during this time. Therefore sin ^^ must decrease in this 

rrr 

interval from — to sin~i ( — 1), which therefore can only 
be — — . Hence if t is the time of a single oscillation, 



4 



■l-l-c-i) 



€ 



or t = IT- 

''9 

This value of t is called the period of vibration. 
General Solution. For large vibrations the above ap- 
proximate solution is not sufficiently accurate. 
Integrating equation (1), 

vdv = — gl sin OdS., 

we obtain ~^ — 9^ ^^^ -\- 0. 

When i; = 0, 6> = B^. 

.'. C = — gl cos 6^ 
and ?;2= 2^Z(cos ^ — cos ^j). 



CURVILINEAR MOTION 217 



.♦. l^= -V2^^ (cos (9 -COS (9,) 

at 

or -7====== - \/^ 6?L 



^^ 



VcOS ^ — COS 0^ ^ I 

Q 

Since cos 6 = 1 — 2 sin^ -, this equation may be written 



dd 



4 



= -2yjS.dt. 



0. • 2 ^ ^ 

^ — sm^ - 

2 2 



If ^j is the time that it takes for the pendulum to descend 
from e = d^to 0= 0, then 

\/sin^-^— sin^ - 
>/ 2 2 

^2 2 

The time of vibration, t^ is double this, or 
^^J7 r''^ c^<9 



-vr 



^Sin2^-sin2- 

Change the variable by the substitution 

.6 • d. . . 
Sin - = sin—i sm <p. 

2 2 

Then - cos - dO = sin -i cos 6 d6, 

2 2 2 ^ ^ 

and - — '^ ^^ 



=2V^r 



\/l — sin^— 1 sin^ <f> 
V 2 



218 APPLIED MECHANICS FOR ENGINEERS 

This integral is an ellipticititegral. The expression can be 
integrated only by expanding into a series and integrating 
term by term. 

Thus, letting k = sin -i, 

TT 

= 2\f^ r ( 1 + -A;2 sin2 ch + ^^ -k^ sin^ 6 
^gJo' 2 ^1.222 ^ 

1.2. 3 23 
From the reduction formula 

sin"^ a;c?x = -\ I sin"*"^ xdx 

n m ^ 

there follows 

IT TT 

P- m J m-in. ^_2 -, _ (m-l)(7^-3)....S.l 7r 
I sm"* xdx = I sin"' ^ a^aa; — ; ^ ] 7. ^ •> 

and hence 

where ^ = sin -^* 
2 

Problem 278. Using the series, compute correct to three figures 
the time of vibration of a simple pendulum of length 3 ft. when the 
jiendulum swings through 20°. Compare with the time obtained 

from t = TT-v'— 



CURVILINEAR MOTION 219 

Problem 279. Compute the time of vibration of a simple pendu- 
lum of length 4 ft. when swinging through an angle of 80^. 

Problem 280. A pendulum vibrates seconds at a certain place and 
at another place it makes 60 more vibrations in 12 hours. Compare 
the values of g at the two places. 

Problem 281. Show that in the simple pendulum for any angle 
of swing the value of the tension in tlie string is 

t=g[' + -'1-'''), 

where h and ij have the meaning given them in Art. 120. 

Hence show that the tension is equal to the weight w^hen the 
weight has descended through one third the original vertical distance. 

Problem 282. The weight of a chandelier is 300 lb., and the dis- 
tance of its center of gravity from the ceiling is 16 ft. Neglecting 
the weight of the supporting chain, find how much the tension in the 
chain will be increased if the chandelier is set swinging through an 
angle of 8^ measured at the ceiling. 

121. Cycloidal Pendulum. — It has been found that a 
pendukim may be obtained whose period of vibration is 
constant by allowing the string to wraj:) itself around a 
cycloid as shown in Fig. 178. The pendulum hangs from 
tlie point A. AB and AC are cych)idal guides around 
which the string wraps as the penduhim swings. This 
causes the length of the pendulum to continually cliange 
and the pendulum '"bob " to move in another cycloidal 
curve COB. The equation of this curve referred to the 
axes X and y is 

a: = ^ vers i-^jA ~^\^^~ .^^• 

In Art. 119 it was seen that v^=2g {h — y) represented 
the velocity of a body moving in a vertical curve when 



220 APPLIED MECHANICS FOR ENGINEERS 



only the force of gravity and a force normal to the path 
of the curve acted. We may make use of the equation in 

r 




Fig. 178 

this case, since the same conditions exist. We may write 

7. y/dx^ + dy^ . ds 

at = — '' '— , since v = — 

V2g{h-i/) dt 

From the equation of the curve, we find 

dx = ^ ~2^ • d^^ 
so that 



2y 






taking the negative sign, since ^ is a decreasing function 
of y. 



Therefore ^=\/ — 



CURVILINEAR MOTION 221 

r 



9 



-1 2 y 
vers ^ — ^ 

h 






The whole time of vibration is twice this value, so that 
the time of vibration 

This expression is independent of A, so that all vibrations 
are made in the same time. The motion is therefore 
isochronal. 

Problem 283. Using the calculus formula for radius of curva- 
ture, 



3 



r = 



(h/Yl^ 



show that the radius of curvature of the cycloid of the above article 
at any point is 



r=V/(/-2?/), 
and hence show that the tension in the string for any position is 

where G is the weight of the pendulum bob. 

Problem 284. Assuming a value of h in terms of Z, say h = - , 

4 

plot the curve representing the tension in terms of //. For what 
position of the pendulum is the tension a maximum ? 
Show that for any swing the maximum tension is 



T=o['-±p) 



Problem 285. A particle slides from rest down an inverted 
cycloid. Prove that its vertical velocity is greatest when it has de- 




222 APPLIED MECHANICS FOR ENGINEERS 

scended through one half the original vertical distance above the 
lowest point of the curve. What is the vertical velocity at that point? 

122. Motion of Projectile in Vacuo. — A method, slightly 
different from the preceding, of dealing with a problem 

of curvilinear 
motion, is il- 
lustrated in 
the present ar- 
ticle. It is de- 
sired to find 
the path taken 
^lo. 179 by a body pro- 

jected with a velocity Vq at an angle of elevation a, when 
the resistance of the air is neglected. (See Fig. 179.) 
Let the point of projection be taken as origin and the 2;-axis 
horizontal. Then, since there is no horizontal force act- 
ing on the body, a^ = 0, so that 

d X /-v 

doc 
and — = constant = v^ cos a, 

therefore x = VqCos a{t). 

In a similar way we know that the vertical acceleration 
ay = — g^ since the only force acting is G. 

Then, g = -, 

and -^ — — qt-\- constant. 

dt 



CURVILINEAR MOTION 223 

This equation may be rewritten 

Vy= — gt -\- constant. 
To determine this constant of integration, we put f = 0, 
ajid Vy = Vq sin a ; 

therefore -^ = — gt -\- v^ sin a 

dt 

and y = — ^gt^ -{- Vq sin a(^). 

Eliminating t between the equations in x and ?/, we get 

y^xtana- f ^" ^ 

2 t^o cos- a 

as the equation of the path of the projectile. This is evi- 
dently a parabola, with its axis vertical. 

Range. To find the range or horizontal distance d we 
put g = ; then x — and x = d^ 



.2. 



J.1 i , VnSm2a 

so that d = — . 

a 

From this it is clear that the greatest range is given when 

a = 45°, since then c? = — . 

The Greatest If eight. The greatest height to which the 

v sm 2 ot 
projectile will rise is found by putting x = -^^ in 

the equation of the curve and solving for g. This gives 

, vl sin^ a 

and the angle that gives the greatest heiglit is a = 90°. 

For this case h = -—. Tliis is the case tliat has already 
2^ 



224 



APPLIED MECHANICS FOR ENGINEERS 



been considered under the head of a body projected verti- 
cally upward. 

Problem 286. A fire hose delivers water with a nozzle velocity vq, 
at an angle of elevation a. How high up on a vertical wall, situated 
at a distance <J' from the nozzle, will the water be thrown V It should 
be said that water thrown from a nozzle in a non-resisting medium 
takes a parabolic path and follows the same laws as projectiles. 

Problem 287- What at least must be the nozzle velocity of water 
thrown upon a burning building, 200 ft. high, the angle of elevation 
of the curve being 60° ? 

Problem 288. The muzzle velocity of a gun is 500 ft. per second. 
Find its greatest range when stationed on the side of a hill which 
makes an angle of 10° with the horizontal : (a) up the hill, (h) down 
the hill. If the hillside is a plane, prove that the area commanded 
by the gun is an ellipse, of which the gun is a focus. 




Fig. 180 



CURVILINEAR MOTION 



225 



Problem 289. From the foot of a plane inclined (3 to the hori- 
zontal a projectile is fired at an angle a to the horizontal up the plane. 
Prove that the range on the inclined plane is 

2 t'^^ cos a sin (« — fS) 
(J cos^ ^ 

Problem 290. Prove that the initial velocity vq is the same as 
that of a body falling freely from the directrix of the parabolic path 
to the point O on the curve. Show that the velocity of the body at 
any point on the curve is the same as would be acquired in falling 
freely from the directrix to that point. (See Fig. 179.) 

Problem 291. A ball whose weight is 64.4 lb., shown in Fig. 180, 
starts from rest at A and rolls without friction in a circular path to 
the point B, where it is 
projected from the cir- 
cular path horizontally. 
Find (a) the velocity 
at B, (b) the equation 
of its path after leaving 
B, and (c) the distance 
d from a vertical 
through B, where it 
strikes a horizontal 10 
ft. below B. 




/ 





I lO =i12 LBS. 



ff 



Problem 292. If 

the body in Problem 
291 had moved along 
a straight line from A 
to 5 and was then pro- 
jected, find, as in the c'l 
preceding problem, (a), j..^^ ^^^ 
(6), and (e). 

Problem 293. A body who.se weight is 12 lb. swings as a circular 
pendulum, as shown in Fig. 181, from A to B^ when the string breaks. 
Q 



226 APPLIED MECHANICS FOR ENGINEERS 

Find (a) the velocity at B, (/v) the equation of its path after leaving 
B, and (c) the distance d where it strikes a horizontal 5 ft. below B. 

Problem 294. The muzzle velocity of a gun situated at a height of 
300 ft. above a horizontal plane is 2000 ft. per second. Find the area 
of plane covered by the gun. 

Problem 295. A projectile is fired from a mortar gun with an 
initial velocity of 25 mi. a minute. What is the maximum range 
on the horizontal, neglecting air resistance? 

123. Motion of Projectile in Resisting Medium. — It was 
found by Rollins and others (see Encyclopaedia Britannica 
— " Gunnery ") that the formula for projectiles in vacuo did 
nothold when the projectile moved in the atmosphere. That 
is, that the path followed by the projectile was not para- 
bolic, but on account of the resistance of the atmosphere 
the range was much less than that given by the parabola. 

A formula constructed by Helie, empirically modifying 

the parabolic formula, is 

, gx^ f\ , kx 

y = X tan a — ^ ^ [ 1 

2 cos^ a\v2 Vq 

where A: = 0.0000000458 ^, c? being the diameter of the 

w 

projectile in inches, and w its weight in pounds. This 
gives the simplest formula for roughly constructing a 
range table. 

Professor Bashforth of Woolrich found, from a series 
of experiments made by him, that for velocities between 
900 and 1100 ft. per second the resistance varied as v^^ for 
velocities between 1100 and 1350 ft. per second the re- 
sistance varied as v^, and for velocities above 1350 ft. per 
second the resistance varied as v^. 



CURVILINEAR MOTION 



227 



In addition to the resistance of the air other factors tend 
to change the path of the projectile from the parabolic 
form, viz. the velocity of the wind and the rotation of the 
projectile itself. Most projectiles are given a right-handed 
rotation, and this causes them to bear away to the right 
upon leaving the gun. This is called drift Correction 
is made for drift and wind velocity upon firing. 

Problem 296. Taking r^ = 1000 f/s, a =45^, d =Q in., w = I.IO lb., 
find the range from H^lie's formula. Plot to the same scale the curve 
followed by the projectile and the parabola it would follow if there 
were no resistance. 

Problem 297. With the projectile of the preceding problem and 
I'^j = 1200 f/s, find the angle of elevation to strike a point 200 ft. high, 
distant 1000 ft. horizontally, (a) using Il^lie's formula, (b) using the 
parabola. 

124. Motion in a Twisted Curve. — -If a particle moves 
along a curve in space, it follows, as in the case of a two- 
dimensional curve, that the ve- 
locity of the particle at any point 
of the curve is in the direction of 
the tangent to the curve at that 
point, and its value is 

ds 



V = 



dt 



(Fig. 182.) 



If the tangent makes angles «, 
/Q, 7, with the coordinate axes, 
then 




dx 

v- = — = V cos a. 

dt 



dt 



Fkj. 182 

dz 

V, = = V COS 7. 

dt 



If a is the acceleration of a particle when at the point P 



228 APPLIED MECHANICS FOR ENGINEERS 

and a^., ^j^, a^, the components of a parallel to the axes, then 

_ (Px _ d{v cos a) 
''""^~ Jt ' 

_ dP'y d(y cos /3) 

_ dP'Z _ d(v cos 7) 
""'"It^" Jt ' 

do . da 

or a^ = cos a - — v sin a -— , 

dt dt 

adv . n dB 

a^ = cos p — V sm S — , 

'' c?^ dt 

dv . c?7 

«^ = cos 7 V sin 7 — ^ • 

c?^ dt 

The sum of the projections of a^., a^, a^ on the tangent 
is the component of the acceleration along the tangent; i.e. 

at = a,, cos (t-{- ay cos ^ -\- a^ cos 7 

= (cos^ a + cos^ /5 + cos^ 7) — 

ai/ 

( ' da ^ . a od/3 , . dy\ 

— vi sm a cos a 1- sm p cos p -^ + sm 7 cos 7 ~ • 

\ 6?^ c?^ dtj 

But cos^ a + cos^ /3 + cos^ 7 = 1, and therefore 

— 2( cos a sin a — + cos fi sin /3 -7- H- cos 7 sin 7 -^ ) = 0. 

\ a^ dt dt/ 

Therefore ««=^!- 

If a particle slides without friction down any space 
curve, it can be shown as in the corresponding case of the 



CURVILINEAR MOTION 



229 



plane curve (Art. 109) that the velocity attained in 
descending a vertical distance h from rest is given by 

v^=2gh. 

In the actual case of a body sliding along a curved 
chute, friction on the side and bottom of the chute and 
air resistance reduce tlie velocity below that given by the 
above formula. 

As an illustration consider the motion of a body along a 
helical chute with vertical axis (Fig. 183). 

z 





Fig. 183 

If the 2-axis is the axis of the helix and the curve passes 
through a point on the rr-axis, then 

X = r cos ^, 
y = r sin ^, 

27r 

where d is the 'pitcli of the helix, or vertical distance 
traveled by the generating point per turn. Developing 
one thread of the helix, it is seen that the tangent makes a 
constant angle 7 with the ai-axis such that 



tan 7 = 



d 



230 APPLIED MECHANICS FOR ENGINEERS 

During the motion of the body the chute exerts a force 

towards the axis of the curve of value • This force 

r 

causes a retarding friction force along the tangent of value 

c , where c is a constant depending upon the roughness 

r 

of surface of chute and body. In addition there is a tan- 
gential friction force due to the force exerted by the bot- 
tom of the chute on the body. The component of the 
weight perpendicular to the bottom of the chute is Wsin 7 
and the friction force caused by it is cWsiny. Therefore 
the total retarding force is 

c[ W sm 7 H 

\ gr 

The component of the weight along the tangent is TFcos 7. 
Hence the net accelerating force along the tangent is 



Fi=W{ cos 7 — c? sin 7 — c ^ 



grj 



Wc 



= — -{h"^ — v^)^ where b^ = '-^ cos 7 — fjr sin 7. 
gr c 

Hence the acceleration along the tangent is 

m r 

or ^ = ^(52_^2). 

at r 

1 heretore — = - at. 

Ir — v^ r 

Integrating, -- log =t+ C\. 

'lb b — V r 



CURVILINEAR MOTION 231 



If the body starts from rest, v = when ^ = 0. 

.-. C' =^locrl = 0. 
^ -lb ° 

i hereiore = e r . 



h — V 



Solving for v, 



2bC f 

■,e>- — 1 

^ = ^-T7 

e r 4-1 



= ^> 1- 



e r +1 

This formula shows that v remains always less than the 
constant quantity b. 

On account of the velocity not exceeding a fixed value 
this form of chute is useful in sending packages from one 
floor of a building to another. 

Problem 298. Given a helical chute of radius 3 ft. and pitch 2 ft., 
find the velocity of a small body descending from rest after 5 sec, 
after 10 sec, if the value of c is .05. 




CHAPTER XI 

ROTARY MOTION 

125. Angular Velocity and Angular Acceleration of a Par- 
ticle. — If a particle moves in any plane curve, the rate at 

which a line joining the particle to 
a fixed point in the plane is turning 
is called the angular velocity of the 
particle with respect to that point. 
Thus, if 6 is the angle which 

the line ioininef the particle makes 
Fig. 184 j &> r 

with a fixed line through (Fig. 
184), the angular velocity, o), of the particle with respect 
to is 

dt 

The rate of change of the angular velocitj^ is called 
the a^igular acceleration of the particle with respect to the 
point. 

Thus, if a is the angular acceleration of the particle 
with respect to (9, 

a = — - or a = , 

dt' dt' 

The angular acceleration may also be written in the form 

d(o dO dta 

CC = — , or a = 0) — . 

dO dt d9' 

232 



ROTARY MOTION 233 

The angular velocity and angular acceleration of a 
particle moving in a given path with a given speed de- 
pend upon the point to which they are referred. 

Problem 299. Prove that if a particle moves around a circle, the 
angular velt>city of the particle with respect to a point on the circum- 
ference is equal at any instant to one half the angular velocity of the 
particle with respect to the center of the circle, and hence that the 
angular velocity of the particle at any instant is the same with re- 
spect to all points on the circumference. 

Show also that a like statement holds for angular acceleration. 

Problem 300. Show that if a particle have constant angular 
acceleration with respect to any point, the following formulae hold : 

t 

where 6 is the angle turned through in the time <, Wq the initial an- 
gular velocity, and w the angular velocity at the instant t. Compare 
these formulae with those of Art. 104. 

Problem 301. A flywheel making 100 revolutions per minute is 
brought to rest in 2 min. Find the angular acceleration a and the 
angle 6 turned through before coming to rest. 

Problem 302. A flywheel is at rest, and it is desired to bring it 
to a velocity of 300 radians per minute in \ min. Find the angular 
acceleration « necessary and the number of revolutions required. 
What is the angular velocity <u at the end of 10 sec? 

126. Motion in a Circle. — If a particle is moving in a 
circle, tlie linear velocity at any instant is 

v=—. (Art. 115.) 

dt 



234 



APPLIED MECHANICS FOR ENGINEERS 



But here s = r6, 

where s is the arc passed over and 6 the angle subtended 

by this arc at the center. 

ds _ dO 

dt dt 




or 



V = Vta. 



Taking the second derivative, 



Fig. 185 



d^^ ^ 
df^ ^ dt^ 



dps 
From Art. 117 — - is the tangen- 
dt^ 



tial component, a^, of the linear 
acceleration of the point, and hence 

at = ra. 

Problem 303. Prove that for a particle moving in a circle with 
angular velocity cd about the center, 



dx 
dt 



= - yo), 



dy 
dt 



where x and ,y are the coordinates of the position of the particle re- 
ferred to rectangular axes through the center. 

Problem 304. The balance wheel of a watch goes backward and 
forward in | sec. The angle through which it turns is 180°. Find 
the greatest angular acceleration and the greatest angular velocity, 
given that the angular acceleration varies directly as the angular 
displacement and is toward the neutral position of the spring. 

Problem 305. Suppose the flywheel in Problem 302 to be G ft. 
in diameter. After arriving at the desired angular velocity, what is 
the tangential velocity of a point on the rim? "What has been the 
tangential acceleration of this point, if constant? 



ROTARY MOTION 



235 




Fig. 186 



Problem 306. Assume that the angular acceleration of a particle 
varies inversely as the square of the angle turned through; find 
the relation between oi and 0, and t and 0. 

127. Angular Velocity with Respect to an Axis. — If a par- 
ticle moves along any space curve, its angular velocity 
with respect to any fixed axis is 
the rate at which a line passing 
tlirough the particle and the axis, 
perpendicular to the axis, is turn- 
ing about the axis ; or it is the 
angular velocity of the projection 
of the moving particle on a plane 
perpendicular to the axis with 
respect to the foot of the axis on that plane (Fig. 186). 

Problem 307, The coordinates of a moving particle are 

a; = r cos ct, y = r sin ct, z = kt, 

where r, c, and k are constants and t is the time. Prove that the 
angular velocity about the c-axis is c. What is the curve traced out ? 

128. Plane Motion of a Body. — A body is said to have 
plane motion when each point of the body moves in a fixed 
plane. The plane in which the center of gravity of the 
body moves will be called the plane of motion. 

The rate of turning of any fixed line of the body in the 
plane of motion is called the angular velocity of the body. 
It should be noted that the idea of rotation about an axis 
does not enter into this definition of angular velocity of a 
body. 

129. Relation between the Velocities of Points of a Body 
having Plane Motion. — Let A and B (Fig. 187) be two 



236 



APPLIED MECHANICS FOR ENGINEERS 




Fig. 187 



points of the body in the 
plane of motion. Choose 
axes OX and Oy in the 
plane of motion. 

If (x^ y) and {x\ y') are 
the coordinates of A and B 
respectively and 6 is the 
angle which AB makes 
with the a^-axis, then 



x' = X -\-r cos ^, 
y' =y ^r sin 6, 

where r is the constant distance AB. 



Therefore 



dx' 



dx 

~di 



r sm 



6 



dd 
dt' 



dy' dy , ^dO 

-^- = -^ -\- r cos u — • 

dt dt dt 



(1) 

(2) 



(3) 
(4) 



If A were a point fixed in space, — - and -^ would be 
^ ^ dt dt 

zero and the components of the velocity of B would 

become 

dx' • ^ dO 

__= — ram 6—-, 

dt dt 

dy' a d6 

-^- = r cos u -— ' 



dt 
dx 



dt 



dy 



If A is in motion, -- and -^ are the components of the 

dt dt 

velocity of A. It therefore follows from equations (3) and 

(4) that the velocity of B is compounded of the velocity of 

A and the velocity of B relative to A regarded as at rest. 



ROTARY MOTION 



237 



Hence if v^ is the velocity of A at any instant and co 
the angular velocity of the body at that instant, the veloc- 
ity of B is the resultant of Va and 
ro), the latter being at right angles -^ 

to AB (Fig. 187). j)^ 

Problem 308. A wheel of radius 3 ft. rolls 
along a straight line, the velocity of the center 
being 10 f/s (Fig. 188). Find the velocities 
of the points A, B, D, and E shown in the 
figure. Find the velocity of A relative to Fig. 188 

B, of E relative to .1, of D relative to E. 

Problem 309. Show that the velocity of B in Fig. 187 is 




Vj = Vi'l + rV^ + 2 ruiVa sin 0. 

130. Relation between the Accelerations of Points of a Body 
having- Plane Motion. — Taking the derivatives of the 
equations (8) and (4) of Art. 129, it follows that the ac- 
celeration of a point in the plane 
of motion of a body having plane 
motion is composed of the accel- 
eration of any other point in the 
plane of motion and the accelera- 
tion of the first point relative to 
tlie second regarded as at rest. 
Thus if a body in plane motion 
has angular velocity o) and angular acceleration a, the accel- 
eration Uf, of a point B is made up of a, the acceleration of 
A, rco^ along BA, and ra perpendicular to AB (Fig. 189). 

Problem 310. Find the acceleration of each of the points A, B, 
D, E, in Problem 308, given that the velocity of the center is con- 
stant. Represent these velocities by vectors. 




Fig. 189 



238 



APPLIED MECHANICS FOR ENGINEERS 



Problem 311. If the point C in Problem 308 has an acceleration 
of 2 f/s- in the direction of motion at the given instant, find the 
acceleration of each of the points A, B, D, E. 

Problem 312. Show that for a wheel rolling along a straight 
line the tangential velocity and acceleration of any point of the cir- 
cumference relative to the center regarded as at rest are the same in 
magnitude as the velocity and acceleration of the center of the wheel. 

Problem 313. A Ic^omotive drive wheel 6 ft. in diameter rolls 
along a level track. Find the greatest tangential acceleration and 
the greatest normal acceleration of any point on the tread, («) when 
the constant velocity v with which the wheel moves along the track 
is 60 mi. per hour, (b) when the engine is slowing down uniformly 
and has a velocity of 30 mi. per hour at the end of 3 min., (c) when 
the engine is starting up uniformly and has a velocity of 30 mi. per 
hour at the end of 5 min. 



131. Instantaneous Axis of Rotation for Plane Motion. — 
Let A and B be two points of a body having plane mo- 
tion, lying in the plane of motion. 
If the body does not have a motion 
of translation only, A and B can 
always be chosen so that their 
directions of motion are not par- 
allel. Through A draw in the 
plane of motion a line, AO^ per- 
pendicular to the direction of 
motions of A. Any point of the 
body on AO, if it is in motion at 
all, must be moving perpendicular to A 0, since any two 
points of the body remain always the same distance apart. 
Likewise, a point on OB, which is perpendicular to the 
direction of motion of B, must, if moving at all, be moving 




Fig. 190 



ROTARY MOTION 



239 



perpendicular to OB. The point of intersection, 0, of 
OA and OB, must therefore be at rest, since it cannot at 
the same time be moving perpendicubir to OA and OB. 

The line through 0, perpendicubir to the plane of mo- 
tion, is called the instantaneous axis of rotation. Every 
point on this axis has velocity zero at the given instant, 
and the motion of the body is at the instant one of rota- 
tion about this line. 

If the figure in motion is a plane figure moving in the 
plane in which it lies, the point is called the instanta- 
neous center of rotation. 

The path traced by the instantaneous center is called 
the eentrode. 

If OA = rj, OB = r^, and the velocities of A and B are 
respectively v^ and v^., then the angular velocity of the 



It follows that fj : v^ z= r^: r^- 

Problem 314. A straight line of length a moves with its ends on two 
axes at right angles to each other. Prove that the eentrode is a quadrant 
of a circle of radius a and center at the intersection of the two axes. 

Problem 315. A connecting rod is 4 ft. long, and the crank-pin 
circle is 1 ft. in radius. Sketch the 
eentrode of the connecting rod for one 
comj)lete stroke. 

Problem 316. The diameter of the 
wheel outlined in Fig. 191 is 6 ft. The 
velocities of the points B and D are as 
indicated by the vectors. Find (r/) the 
instantaneous center, {h) the magnitude Fig. TJl 




240 



APPLIED MECHANICS FOR ENGINEERS 



of the velocity of tlie point D, (c) the angular velocity of the wheel, 
(c/) the velocity of the point E, (e) the velocity of slipping of the 
point A of the wheel. 

132. Simultaneous Rotation of a Body about Two Intersect- 
ing Axes. — Suppose a body rotating about an axis OA 
(Fig. 192) with an angular velocity &)j, relative to a frame- 
work attached to 
OA, and let OA 
and the attached 
framework be at the 
same instant rotat- 
ing about an axis 
OB witii angular 
velocity oo^. 

If P be a point of 
the body, or a point of space thought of as moving with 
the body, in the plane containing OA and OB, between 
OA and OB, and distant r^ and rg, respectively, from these 
lines, the velocity of P due to the two rotations is 




Fig. 192 



V = r.oD. — roft)o. 



!«/! 



2"'2- 



If P is so chosen that r^ and rg satisfy the relation r-^ : r^ 
= ft>2 : Wj, the velocity of P is zero. The locus of all such 
points is clearly a straight line through 0. Hence there 
is a straight line, OP, in the plane of the two intersecting 
axes of rotation wliich is at rest at the given instant, and 
about which the body is rotating at that instant. 

To find the angular velocity, co, of the body about this 
instantaneous axis OP, consider the motion of the point 



ROTARY MOTION 



1^41 



on OA. Draw CU perpendicular to OB and OF per- 
pendicular to OP. 

The point C has angular velocit}^ 0)2 about OB, and 
hence its velocity is UCco^. Its angular velocity about 
OP is o) and hence its velocity is FCco. 

.-. FCco=EC(o^. 



Now 



jFe= 0(7 sin a and ^0= 0(7 sin (9. 



(o sin « = ftjg sin ^. 



a) 



Again, ^(7 = rj cos a, and EC =r^-\-r^ cos ^. 

Therefore r^co cos a = ^20^2 + t^w.^ cos 6, 

But 

and hence 



^2®2 = ^\^v 



CO cos a = ft)j + 0)2 COS 6. 

Squaring and adding equations (1) and (2), 
«■- = 0)2 + «2 -I- 2 «i«2 cos 6. 



Dividing (1) by (2), 

tan a = 



(1)2 sin ^ 
Wj + cOg cos ^ 



(2) 



(3) 



(4) 



Equations (3) and (4) are exactly the equations that 
would be obtained by re- 
garding Q)j and 0)2 as vec- 
tors and finding their re- 
sultant as in Fig. 193. 

Hence we may represent 
angular velocity of a body 
about a line by a vector 
laid off on that line, equal 
in length to the numerical 
value of the angular ve- 




242 



APPLIED MECHANICS FOR ENGINEERS 



locity, and may combine simultaneous angular velocities 
about two intersecting lines by combining their vectors in 
the usual way. The resultant vector represents the value 
of the resultant angular velocity and coincides with the 
instantaneous axis of rotation of the body. 

The vector representing an angular velocity is made 
to point opposite to the direction from which the rota- 
tion appears counter-clockwise to an observer looking 
along the axis of rotation. 

It is evident at once, then, that the angular velocity 
of a body about a line may be regarded as made up of 

component angular velocities about 
two or more axes intersecting that 
line, obtained by using the parallelo- 
gram law of vectors. In particular 
a rotation about an axis may be re- 
garded as made up of the simulta- 
neous rotations about three axes at 
/ ^^ I right angles to each other and inter- 

secting the given line (Fig. 194). 
If the angular velocity of the body 
about the line is «, the angles which this line makes with 
the three rectangular axes, a, /S, 7, and the angular velocities 
of the body about these axes w^, (Oy, &>, respectively, then 

co^ = 0) cos a, (Oy= CO cos yS, o)^= co cos 7, 

and ft)2 = ft);. + ft)2 + ft)2. 

Problem 317. Prove that the vector representation holds for 
simultaneous rotation about two parallel axes, i.e. that the resultant 
angular velocity of Wj and o)„ is 




Fig. 194 



ROTARY MOTION 



243 



and that the instantaueous axis of rotation divides the line joining 
the axes of Wj and ojo in the inverse ratio of o)^ and w.,. (See Fig. 195.) 

Problem 318. A wheel is making 120 r. p. m. about a shaft and 
the shaft at the same time is making 90 r. p. m. about a line perpen- 
dicular to the axis of the shaft. Find the instantaneous axis of the 
shaft at any instant. 



■r^-^ 



-fl — ^^^ — >- 



■0: 



i? 



C 



VcJo 



D 



■ff^ 



Fig. 195 



Fio. lOfi 



Problem 319. A body has simultaneous rotations (o^= 1 tt rad./sec, 
W2 = 3 TT rad./sec, about the parallel lines of Fig. 196 in the direc- 
tions indicated. Locate the instantaneous axis of rotation and find 
the angular velocity of the body about that axis. 

133. Rotation of the Earth. Foucault's Pendulum. — T.et 
(o be the angular velocity of the 
earth about its axis of rotation 
CF (Fig. 197). This angular 
velocity can be resolved into two 
components about rectangular 
axes CA and CB. If X is tlie 
latitude of the position of A^ the 
component of o) about CA is (o 
sin X. If a plane containing CA 
could be fixed so as not to turn 

about CA, tlie rotation of the earth about CA would be 
indicated by the apparent rotation of the plane in the 




Fkj. liiT 



244 APPLIED MECHANICS FOR ENGINEERS 

opposite direction about (7A, since our directions are 
measured with reference to the earth's surface. 

Such a phine is the plane of vibration of a heavy weight 
suspended by a long wire. 

Foucault was the first to demonstrate the rotation of 
the earth by this method. 

Problem 320. Show that the rotation of the earth about a di- 
ameter through a point on the equator is zero. 

Problem 321. Show that the time for the plane of the Foucault'.s 

pendulum to turn through 360^ is — — hours, where A is the latitude 

sin A 
of the place. 

Problem 322. The plane of a Foucault's pendulum is observed 
to turn through 26^ in 2 hr. 20 min. Find the latitude of the place. 



CHAPTER XII 



WORK AND ENERGY 

134. Definitions. — If a force, constant in niacfnitude 
and direction, acts at a fixed point of a body, the force is 
said to do work on the body wlien the point of application 
of the force has a displacement with a component in the 
direction in which the force acts. 

The product of the force and the component of the dis- 
placement in the direction of the force is defined to be 
the work done on the body by the force in that displace- 
ment. If there is a displacement opposite to the direction 
of the force, work is said to be done against the force. 

For example, in Fig. 198, let the forces P, 72, (r, iVact 
upon the block as the block moves from the position A to 




b'm. i\)S 

the position B. If 5(7 is perpendicular to the line of P, 
the displacement in tlie direction of P is AC and tlie 
work done by P on the block is P - AC. If AB = s, then 
AC=8 cos 6, and llu? work done by P is Ps cos 0. Tliis 

245 



246 APPLIED MECHANICS FOR ENGINEERS 

may be written P cos 6 - s. But P cos 6 is the component 
of P in the direction of the displacement of the point of 
application, and therefore the work done by P is equal 
to the product of the displacement of the point of appli- 
cation and the component of the force in the direction of 
the displacement. 

Since iV and Gr are at right angles to the displacement, 
no work is done by iVor Gr. 

Since the displacement is opposite to the direction of i?, 
work is done against i2, the amount being Rs. 

If the body had moved from P to ^ during the action 
of the given forces, the work done by M would be Ps, and 
the work done against P would be P cos -s . 

If the point of application changes in the body, the 
work done on the body is the product of the force and 
the displacement, the displacement meaning the actual 
displacement of the force in space plus the relative dis- 
placement of the moving point of application past the 
initial point of the force vector in the direction opposite to 
the direction in which the force acts. If this relative dis- 
placement is in the direction in which the force acts, it must 
be subtracted from the displacement of the force in space. 

For example, in 

jj Fig. 199 suppose a 

force P, applied to 

the block P, drags P 

along" A and at the 



'^ 



& - 



^ 






> 



Fig. 199 

aion^ 

same time moves A. Let P be the forward force exerted 
by P on A. If when A moves forward a distance «, B 
moves relative to ^ a distance s', then the relative dis- 



WORK AND ENERGY 247 

placement of the point of application past the initial point 
of the force is s' in the direction in wliich the force acts, 
and hence the work done by F on A is F(^8-\-s' — s') or 
Fs. On the other hand, a force F acts backward on B and 
the block B does work against this resistance to an amount 
equal to ^ (s + s'). Hence the total work done against 
the forces of resistance between the bodies is Fs\ or the 
product of the force and the distance one body moves rel- 
ative to the other. 

135. Friction Forces. — Wherever the surfaces of two 
bodies come in contact and there is motion of one body 
along the other, or any force tending to produce such 
motion, each body exerts upon the other a force along 
their common surface, a force tending to prevent the rela- 
tive motion of one surface along the other. This resist- 
ing force is known as friction. The laws of friction are 
discussed in a later chapter. In general the law of fric- 
tion of unlubricated surfaces may be expressed by saying 
that the force of friction for two given surfaces is pro- 
portional to the normal pressure between the surfaces. 

The ratio of the force of friction to the 
normal force is called the coefficient of 
friction. 

Consider the work done ao'ainst fric- 
tion on an axle rotating in a fixed bear- 
ing (Fig. 200). Here tlie point of ap- ^iq. -200 
plication of the friction force, F, changes 
in the body while the initial point of the force vector re- 
mains fixed in space. The displacement is then the dis- 





248 APPLIED MECHANICS FOR ENGINEERS 

placement of the outer points of the axle past the fixed 
point at which F acts. The work done on the axle by 
friction, in resisting its motion, is therefore the product of 
jP and the distance through which a point on the circum- 
ference moves; or in turning 
through an angle d the work done 
is Frd. 

If the bearing is also in motion, 

as in Fig. 201, then if the bearing 

moves forward s ft. while the 

axle rubs a distance s' ft. past 

Fig'201 ^^^® bearing, the work of F on the 

bearing is Fs, and the work done 

against F on the axle is F(s-\-s'), and hence the total 

work done against friction is Fs' . 

136. Units of Work. — The unit of work involves the 
unit of space and the unit of force. If the distance is 
measured in feet and the force in pounds, the unit of work 
is the work done by a force of 1 lb. when the displacement 
in the direction in which the force acts is 1 ft. 

This unit of work is called the foot-pound, designated 
hy ft.-lb. 

In the c. g. s. system the ei-g is defined as the work 
done by a force of 1 dyne when the displacement in the 
direction in which the force acts is 1 cm. 

The gram- centimeter is the work done by a force of 1 
gram working through a distance of 1 cm., etc. 

137. Work of Components of a Force. — Let the constant 
force P (Fig- 202) acting on a body have the displacement 



WORK AND ENERGY 



249 




Fig. 202 



ABj or 8. Resolve P into any two rectangular com- 
ponents, X and Y. In the given movement the displace- 
ment of -X" is AC, or x, and 
of Y is AD, or y. The work 
done by P in the given dis- 
placement is P cos 6 ' 8. 

But P cos 6 is the projec- 
tion of P on AB and is there- 
fore equal to the sum of the 
projections of X and Y on 
AB; i.e. 

P cos 6 = X cos a + !F sin a. 
.-. P cos 6 ' s = Xs cos « -|- Ts sin a 

Therefore, ^Ag ?^;orA: do7ie hy a constant force in any dis- 
placement is equivalent to the work done by any rectangular 
components of the force in the same displacemeyit. 

Problem 323. A weight of 20 11). is dragged 50 ft. up a plane 
inclined 30^ to the horizon by a constant force, P = 25 lb., acting 
at an angle of 15"^ to the plane. There is a retarding force, R = 5 lb., 
along the plane. Compute the work done by, or against, each force 
acting on the body. Find the work done by the horizontal and 
vertical components of P in the given displacement, and also the work 
done by the components of W perpendicular to and parallel to the 
given plane. 

If the body started from rest, what velocity does it have at the end 
of the 50 ft. ? 

Problem 324. A block, yl, weighing .50 lb., is pulle<l i)y a hori- 
zontal force of 32 11). along a horizontal plane. A block, B, weighing 
20 lb., rests on top of A. If the coefficient of friction between .1 and 



250 APPLIED MECHANICS FOR ENGINEERS 

the plane is ^ and that between B and A is |, will B slip on A ? If 
so, how far would B slip back on A when ^ has moved forward 10 ft. ? 
What work has been done against friction in this motion ? 

Problem 325. In the preceding problem what would the co- 
efficient of friction between yi and B have to be in order that B 
would just not slip? What then would be the total work done 
against friction when A has moved forward 10 ft. ? 

138. Work of a Variable Force. — If a force acting on a 
fixed point of a body changes in magnitude and direction 
and the point on which it acts moves along a curve, the 

,r 





Fig. 203 



work done in any displacement from A to B (Fig. 203) is 
defined to be 

Wp= I P cos (j)ds, 

the integration being taken from A to B^ where (f) is the 
angle at any point of the curve between the force and the 
tangent to the curve. 

Since ds^ dx, and di/ are related just as s, x, and y, in 
Art. 137, we may write 

Wp^fxdx-^frdi/, 

or, if the curve is in three dimensions, 

Wp = fxdx + f Ydy + ^Zdz, 



WORK AND ENERGY 



251 



Problem 326. Prove by iutegration that the work done against 
gravity in bringing the, weight Whom the position A to the position B 
(Fig. 204) along the quadrant of a circle is Wr. 

Problem 327. Show from the definition, 



Work = I P cos cfids, 



that when a weight W is brought from any 
position along any curve to a position h ft. 
higher, the work done against the earth's at- 
traction for the body is W7i. 




Fig. 204 




139. Graphical Representation of Work. — Work has been 
defined as the product of a force and a distance. If the 
force be uniform and equal to P, and the body upon which 
it acts be moved through a distance a, the graphical repre- 
sentation of the work done by P is given 
by the area of a rectangle (Fig. 205) 
constructed on P and a as sides, since 

W= Pa. 

If the force P varies as the distance 
through which the body is moved along 

its line of action, we may represent the work by the area 

of the triangle as shown in Fig. 

206. Let the force be zero 

when the motion begins, and 

let it be Pj when the distance 

passed over along its line of 

action is OA. Then since the 

force varies as the distance, it 

is equal to P as shown in Fig. 

206 for any intermediate distance OC. The total work 





/I 


"D 




r. 




r 






c 


A 



SPACE 

Fig. 206 



252 



APPLIED MECHANICS FOR ENGINEERS 



done, then, in moving the body through a distance OA by 

the variable force P, which varies as the distance, is equal 

(P OA) 
to- — 1- — -' It is seen that this is the same as the work 



done by the average force — ^ acting through the distance 

OA. The resistance of a helical spring varies with the 
elongation or compression. The same law of variation 
holds for all elastic bodies. 

Another variation of force with distance with which the 
engineer is frequently concerned, is the case where the 
force varies inversely/ as the distance through which the body 
is moved. In this case, if P is the force and S the distance, 
the relation between force and distance may be expressed, 

PS = const. 

But this represents the equilateral hyperbola. This will be 
made clearer by reference to the specific example of the ex- 
pansion of steam in a steam cylinder. (See Fig. 207.) Up 

to the point of 
cut-off (7, the 
steam pressure 
is the same as 
that in the boiler 
(practically), 
and is constant 
while the piston 
moves from to 
O. At this point, 
the entering 
steam is cut off and the work done must be done by the 




WORK AND ENERGY 



253 



expansion of the steam now in the cylinder. According to 
Mariotte's Law, the pressure varies inversely with the 
volume of steam ; but since the cross section of the cylinder 
is constant, we may say that the pressure varies inversely 
as the distance. From the definition of work, 



Wj. = CPds, 




Fig. 208 



it follows that the area under the curve represents the work 
done. 

The curve ob- 
tained in practice 
representing the re- 
lation between the 
force and distance is 
shown in Fig. 208. 

The curve after 
cut-off is not a true 
hyperbola, and its area is determined by means of a plani- 
meter or by Simpson's Rule. 

140. Power. — The idea of work is independent of time. 
But for economical reasons it is necessary to take into 
consideration this element of time. We must know 
whether certain work has been done in an hour or ten 
hours. For such information a unit of the rate at which 
work is done has been adopted. Tliis unit is called power. 
Power is the rate of doing work. It is the ratio of the work 
done to the time spent in doing that work. 

The unit of power is the horse poiver. This has been 
taken as 550 ft. -lb. per second, or 3tS,000 ft. -lb. per minute. 



254 APPLIED MECHANICS FOR ENGINEERS 

Originally the idea of the rate of work was connected 
with the rate at which a good draft horse could do work. 
This value as used by Watt was 550 ft.-lb. per second. 
The horse power of a steam engine is mean effective 
pressure times distance traveled by the piston per second, 
divided by 550. 

141. Energy. — Energy is the capacity for doing work; 
it is stored-up work. Bodies that are capable of doing 
work due to their position are said to possess potential 
energy. Bodies that are capable of doing work due to 
their motion are said to possess kinetic energy. A familiar 
example of potential energy is the energy possessed by a 
brick as it is in position on the top of a chimney. If the 
brick should fall, its energy at any instant would be 
called kinetic. When the brick strikes the ground, work 
is done in deforming the ground and brick, or perhaps 
even breaking the brick and even generating heat. The 
work done by the brick when it strikes is sufficient to 
use up all the energy that the brick had when it struck. 

142. Conservation of Energy. — The kinetic energy of the 
brick spoken of in the last article was used up in doing work 
on the ground and air, and upon the brick itself, so that the 
kinetic energy that the brick possessed when it struck was 
used up. It was not, however, destroyed, but was trans- 
ferred to other bodies, or into heat. Such transference is 
in accord with the well-known principle of the conservation 
of energy. This principle may be stated as follows : energy 
cannot he created or destroyed. The amount of energy in 
the universe is constant. This means that the energy given 



WORK AND ENERGY 



255 



V=Vi 



up by one body or system of bodies is transferred to some 
other body or bodies. It may be that the energy changes 
its form into light, heat, or electrical energy. 

Energy cannot be created or destroyed ; it is, therefore, 
evident that such a thing as perpetual motion is impossible. 
Such a motion would involve the getting of just a little more 
energy from a system of bodies than was put into them. 

143. Relation between Work and Energy. — Suppose the 
particle of mass il^T acted upon by 
any set of forces to move along 
the path from ^ to ^ (Fig. 209). 
Resolve the forces along and per- 
pendicular to the curve at each 
point. Let Ft be the tangential 
component of the resultant of all 
forces acting on the particle. If 
at is the tangential component of 
the acceleration, then 

Ft = M-at. 

But a, = , = 

dt 

.-. Ftds= Mv • dv. 
The work done on the particle in going from A to -B is 




Fig. 209 



dv ds _ dv 
ds dt ds 



(Art. 117.) 



£ 



Ftds. (Art. 138.) 

If the velocity of the particle is Vq at A and v^ at B, 

. •. CFtds = p Mvdv = l- M(v\ - v2), 
or, The work done on the particle = J, 3f(v2_ ^2^ (\^ 



256 APPLIED MECHANICS FOR ENGINEERS 

The kinetic energy of the particle has been defined as 
the work the particle can do due to its velocity. Suppose 
in Fig. 209 Ff, is opposite to the direction of motion and 
the point B is such that the particle comes to rest there 
from the velocity Vq at A. Then, from the definition, 

K. E. of particle at ^ = I F^ds. 

In this case a^ is negative, a^ = —v—-. 

as 

Therefore Fi = — Mv --, 

as 

and CFids = - C Mvdv = I Mv^. 

Therefore the kinetic energy of a particle with velocity 
V is ^ Mv"^. Equation (1) of this article may then be ex- 
pressed, The work done on a particle in any displacement 
hy the resultant of all the forces acting on the particle is 
equal to the gain in kinetic energy of the particle. 

For any body having a motion of translation only the 
proof applies as well as to a particle. 

Problem 328. A car whose weight is 20 tons, having a velocity 
of 60 mi. per hour, is brought to rest by means of brake friction 
after the power has been shut off. If the tangential force of friction 
of 200 lb. acts on each of the 8 wheels, how far will the car go before 
coming to rest? 

Problem 329. Suppose the car in the preceding problem to be 
moving at the rate of 60 mi. per hour when the power is shut off, 
what tangential force on each of the 8 wheels will bring the car to 
rest in one half a mile? 



WORK AND ENERGY 257 

Problem 330. "What is the kinetic energy of a river 200 ft. wide 
and 15 ft. deep, if it flows at the rate of 1 mi. per hour, the weight of 
a cubic foot of water being 62.5 lb. ? AMiat horse power might be 
developed by using all the water in the river? 

Problem 331. The height of free fall in Niagara Falls is 165 ft. 
Assuming the velocity of the water at the top to be zero, what is its 
velocity at the foot? The flow is approximately 270,000 cu. ft. per 
second. How much kinetic energy is developed per second? What 
H. P. could be developed by using all the water? Counting the height 
of the fall, including rapids above and below, as 216 ft., what H. P. 
could be developed by using all the water? 

Note. It is estimated that the total horse power of Niagara Falls, 
considering the fall as 216 ft., is 7,500,000. The Niagara Falls Power 
Company diverts a part of the volume of water above the rapids into 
their power plants, where it passes through a tunnel into the river 
below the falls. The turbines are 1-40 ft. below the water level, and 
each one is acted upon by a column of water 7 ft. in diameter. The 
estimated power utilized in this way is 220,000 horse power. The 
student should estimate the horse power of each turbine, assuming 
the water to fall from rest through 140 ft. For a full account of the 
power at Niagara Falls, the student is referred to Proc. Inst. C. E., 
Vol. CXXIV, p. 223. 

Problem 332. A body whose weight is 32.2 lb. is pulled up a 
plane, inclined at an angle of 30° with the horizontal, by a horizontal 
force of 250 lb. The motion is resisted by a constant force of friction 
of 10 lb. acting along the plane. If it starts from rest, what will be 
its velocity after it has gone up a distance of 100 ft.? 

Problem 333. The same body as that in the preceding problem is 
projected down the plane with a velocity of 5 ft. per second. How far 
will it go before coming to rest? 

Problem 334. Solve Problem 240 by work and energy. 

Problem 335. Solve Problem 242 by work and energy. 

Problem 336. Solve Problem 243 by work and energy. 

B 



258 



APPLIED MECHANICS FOR ENGINEERS 



Problem 337. A body whose weight is 64.4 lb. falls freely from 
rest from a height of 5 ft. upon a 200-lb. helical spring. Find the 
compression in the spring. 

Problem 338. A weight of 500 lb. is to fall freely from rest 
through a distance of 6 ft. The kinetic energy is to be absorbed 
by alielical spring. Specifications require that the spring shall not be 
compressed more than 2 in. Find the strength of the spring required. 

Problem 339. Specifications state that it shall require 32,000 lb. 
to compress a helical spring 1^ in. What weight falling freely from 
rest through a height of 10 ft. will compress it one inch ? 

Problem 340. The draft rigging of a freight car shown in Fig. 

210 is provided with two 
helical springs, one inside 
the other. The outside 
spring is a 10,000-lb. 
spring, and the inside a 
5000-lb. spring. A car 
weighing 60,000 lb. is pro- 
vided with such a draft 
rigging. While going at 

the rate of 1 mi. per hour it collides with a bumping post. How 

much will the springs be compressed? 

Problem 341. The draft rigging in the preceding problem is 
attached to the first car of a freight train, consisting of 30 cars, each 
weighing 60,000 lb. How much will the springs of the first car be 
elongated if there is 10 lb. pull for each ton of weight when the speed 
is 40 mi. per hour ? The speed is increased to 45 mi. per liour. 
How much w\\\ the springs be elongated if the resistance per ton at 
this speed is 12 lb.? 

Problem 342. The Mallet compound locomotive (Railway Age, 
Aug. 9, 1907) is capable of exerting a draw-bar j)ull of 94,800 lb. 
According to the preceding problem, how many 60,000-lb. cars can be 
pulled at 45 mi. per hour? What strength of spring would be 
necessary for the first car, if the allowable compression is 1^ in. ? 




Fig. 210 



]VOnK AND ENERGY 



'2'y\^ 



Problem 343. An automol>ile going at the speed of 30 mi. per 
lionr comes to the foot of a lull. The power is then shut ofl and the 
machine allowed to "coast" up hill. If the slope of the hill is 1 ft. 
in 50, how far up the hill \vill it go, if friction acting down the plane 
is .06 G', where G is the weight of the machine? 

144. Pile Driver. — A pile driver consists essentially of 
{I hammer of weight G so mounted that it may have a free 
fall from rest upon the pile (Fig. 211). The safe load 
to be placed upon a pile after it has been 
driven is the problem that interests the 
engineer. This is usually determined b}^ 
driving the pile until it sinks only a certain 
fraction of an inch under each blow, then 
the safe load is a fraction of the resistance 
offered by the earth to these last blows. 
The resistance is usually small when the pile 
begins to penetrate the earth, but increases 
as penetration proceeds, until finally, due 
to the last blows, it is nearly constant. 
If we regard the hammer (r as a freely 
falling body, and consider the hammer 
and pile as rigid bodies, and further 
assume, as is usually done, that R for the 
last few blows is constant, we may write 
the work-energy equation. 



]■ 



\] 



t. 



Fig. 211 






since the final kinetic energy is zero and the weight of the 
hammer as a working force is negligible. Tlie distance 8^ 
is the amount of penetration of the pile for the blow in 



260 APPLIED MECHANICS FOR ENGINEERS 

question. But v^ = 2^A, so that 

i Mvi = ah. 

We have then as the value for the supporting power of 
a pile, 

11 = ^. 

A safe value, M' from ^ to |^ of i2, is taken as the safe sup- 
porting power of piles. The factor of safety and the value 
of S;^ for the last blow are usually matters of specification 
in any particular work. This is the formula given by 
Weisbach and Molesworth. Other authorities give 
formulse as follows : 

Trautwine, R = 60 (r VA, if s^ is small, 

5 a</h 



and M = 



Si+1 ' 



1 ^ C h 

Wellington, M = — — - , where h is in feet and s. in inches ; 
»! + 1 

Mc Alpine, J? = 80 [ (^ -h (. 228 VA - 1) 2240] ; 
Goodrich, R = 

For other formulae and a general discussion of the sub- 
ject of the bearing power of piles, the reader is referred to 
Trans. Am. Soc. Civ. Eng., Vol. 48, p. 180. 

The great number of formuke for the supporting power 
of piles is due to the various assumptions that are made 
in deriving them. In deriving the Molesworth formula, 
the hammer and pile were considered as rigid bodies. It 
will be seen that the hammer and pile are both elastic 



WORK AND ENERGY 261 

bodies, both are compressed by the blow; there is friction 
of the hammer with the guides, and the cable attached to 
the hammer runs back over a hoisting drum. There is, 
in most cases, a loss of energy due to brooming of the head 
of the pile. This broomed portion must be cut off before 
noting the penetration due to the last few blows. 

Results of tests have also been taken into consideration 
and have modified the formulae. The Wellington formula 
differs from the Molesworth formula only in the denomi- 
nator, where s^ -f- 1 is used instead of Sy This has been 
done to guard against the very large values of R given 
when Sj is very small. 

Engineers have come to believe that it will be extremely 
difficult to get a general formula that will give very exact 
information as to the bearing power of piles, since soil con- 
ditions are so varied. The more simple formulae with a 
proper factor of safety are used. 

As an illustration of the use of the formula, let us con- 
sider the problem of providing a pile to support 75 tons. 
If the weight of the hammer is 3000 lb., and the height of 
fall 15 ft., the pile will be considered down when 

ah 3000 X 15 Of, oa :„ 
«i = = = .6 it. = o.D in. 

^ B 150,000 
Using a factor of safety of 6, we have Sj = .6 in. 

Problem 344. Compute the value of s^ for the pile in the above 
illustration by using the various formulae given in this article. Com- 
pare the results. 

Problem 345. A pile is driven by a 4000-lb. hammer falling freely 
20 ft. What will be the safe load that the pile will carry if at the last 



262 



APPLIED MECHANICS FOR ENGINEERS 



blow the amount of penetration was J in.? Use a factor of safety of 4. 
Compute by the Molesworth and the Wellington formulae, and 
compare. 

Problem 346. A pile was driven by a steam hammer. The last 
twenty blows showed a penetration of one inch. If two blows of the 
steam hammer cause the same penetration as one blow from a2000-lb. 
hammer falling 20 ft., what weight in tons will the pile support? 
Assume the i)enetration for each of the last few blows the same. 

145. Steam Hammer. — The steam hammer consists es- 
sentially of a steam cylinder mounted vertically and having 

a weight or hammer attached to one 
.^-. end of the piston rod. Let AB 
k i^^^g' 212) be the steam cylinder, 

]_ B tlie piston, and ^the anvil, upon 
which a piece of metal is shown un- 
der the hammer G. The steam 
D ,-i-U „ pressure in the cylinder is constant 

and equal to P, while the piston 
jj passes over a distance a to cut-off, 

and varies inversely as the volume 
over the remaining distance b. Gr, 
the weight of the hammer and 
piston, is also a working force. 
I d The resistance of the 
metal varies during any 



F 



p 




Fig. 212 



sion is greatest. 



blow with the amount of compres- 

. sion. It is zero just as the hammer 

touches the metal, and increases up 

to a maximum when the compres- 

Let R' be the average resistance of the 



WORK AND ENERGY 263 

metal, and M^ the exhaust pressure. Then the work-energy 
equation for the hammer when it lias moved a distance s 
becomes 

M^^M+B, f\u = P r.h + f'P'ds + G f\u, 
or — — ^ -h /ijS = 7^r/ + Cr.s* -f- I J^'ds. 

Since P' varies inversely as the volume of the cylinder, 

., rn const. c 
we may write, J^ = — = - • 

Then the work-energy equation gives 

^ + R,s = Pa + Gs + e logv'- 
^ a 

The term — — ^ is zero, since the motion has been consid- 
ered from rest at the top of the cylinder to a distance s. 
The quantity c may be computed by reading from the in- 
dicator card the value of P' at s, or computed from c — Pa. 
It will be seen that s has been taken greater than a ; that is, 
the piston is beyond the point of cut-off. 

When the hammer finally comes to the face of the metal, 
the work-energy equation may be written 

^ 4- n^h' = Fa + ah' + c log,-, 
z a 

where the distance h' represents the value of s when the 
hammer just touches the metal, and v^ is corresponding 
value of V. This equation gives the kinetic energy of the 
hammer when it strikes the metal. The work-energy 



264 APPLIED MECHANICS FOR ENGINEERS 

equation for the hammer during the compression of the 
piece may now be written 

2 

where d is the amount of compression of the metal due to 
the blow. This is shown by the small figure to the right, 
where the piece of metal has been drawn to a somewhat 
larger scale. 

After the hammer strikes the metal, the steam pressure 
and the weight of the hammer as working forces, and the 
exhaust pressure as a resisting force, have been neglected. 
The work done by these pressures is small, since the dis- 
tance d is small. Approximately, then, the work done on 
the metal equals the kinetic energy at the time of first 
contact. 

Instead of using the value P' = -, and computing the 

s 

integral i P'ds as indicated in the formulae, values of P' 

and s might be read from the indicator diagram (Art. 139) 
and added by means of Simpson's formula (Art. 40). 

As an illustration of the foregoing, let us suppose the 
steam cylinder 25 in. long and 14 in. in diameter; the 
steam pressure P = 18,000 lb. ; the exhaust pressure 
i2i = 2300 lb.; a = 1.2 in.; d=l in.; (?=644 lb.; 
c= 10,800; //=24in. Substituting in the work-energy 
equation, we have for the kinetic energy of the hammer at 
the time of striking the iron, 

. ^^ = 20,490 ft. -lb. 

2 



WORK AND ENERGY 265 

This gives a value for v^ = 45.3 ft. per second as compared 
with 11.3 ft. per second for the same weight freely falling 
through the same distance. 

Investigating now the resistance of the metal, we have, 
under the assumption already made, 

i2'c^ = 20,490 ft.-lb., 
so that i2'= 983,500 lb. 

In the above discussion we have neglected the compres- 
sion of the anvil and hammer due to the blow, and also 
the friction of the piston. 

Problem 347. Find the kinetic energy of the hammer when 
h' = 18 in. Find also v and R', using the same value of d. 

Problem 348. A steam hammer exactly similar to the one given 
in the illustration above is used with a different steam pressure. It is 
only necessary for the work for which it is intended, that the kinetic 
energy of the hammer for a stroke of 2 ft. be 10,000 ft.-lb. 

If a = 7.2 in., R^ = 1800 lb., find what steam pressure P is 
necessary? (Notice that c has a different value from that given 
above.) 

Problem 349. Compute the kinetic energy and velocity of the 
hammer in the illustration (G = 644 lb.) when the piston has moved 
the full length of the cylinder (h' = 25 in.). Assume that there is 
nothing on the anvil. 

Problem 350. What value of h' in the above problem would give 
the hammer the same velocity as it would have if it fell freely from 
rest through the height h ? Compute the kinetic energy for this 
velocity. 

Problem 351. In the illustration given above, what would be 
the value of R' if the steam pressure and G be included as working 
forces, and R^ as a resisting force, during the compression of the 
piece ? 




266 APPLIED MECHANICS FOR ENGINEERS 

Problem 352. In the illustration given above, suppose that, in 
addition to the compression of the piece, \ in., the anvil is compressed 
.02 in. Find the value of IV . 

146. Rotation of a Body about a Fixed Axis. — For a body 
rotating about a fixed axis, if the angle turned through is 
^, tlie angular velocity at any instant is defined as 

0) = > 

dt 

and the angular acceleration as 

dot 

a = 9 

dt 

which may also be written 

Fig. 213 °- " ^p' 

1 d<a dQ dm 

ana a- = = w — • 

For a particle describing a circle of radius r, if v is the 
velocity of the particle at any time and a^ the tangential 
component of the acceleration, then (Fig. 213), 

ds dO 
V = — = r — = rtay 

dt dt 

1 ^ dv d(o 

and at z= -— = r —-= ra, 

dt dt 

147. Work Done by Impressed Forces on a Rotating Body. — 
In Fig. 214 is represented a thin slab of a body rotating 
about a fixed axis AB^ made by planes perpendicular to 
AB. Let the body be acted upon by any set of external 
or impressed forces, of which Pj, P^, ••• are the components 



WORK AND ENERGY 



26" 




ill planes perpendicular to AB^ the other components, not 
shown, being parallel to AB. 

Let dMhe the mass of an clement of the body distant ?• 
from AB. .Vt any instant 
dM has an acceleration due 
to the forces acting on it. 
All the forces acting on dM 
may be resolved into three 
components respectively par- 
allel to AB^ along the radius, 
dJV^ and along the tangent, 
dT. The forces d]^ and dT 
are called the effective forces. y\g '>u 

They are the forces which 

acting on c?iff separated from the body would give it the 
same motion that it has as part of the body. 

If a^ is the tangential acceleration of dM and a is the 
angular acceleration of the body, then 

dT=dMat = radM. 

The forces dN and dT 'dve the components of the result- 
ant of all forces acting on d3I. The forces are made up 
of the action of forces exerted by adjacent particles on dM 
and any external forces that act on dM. The forces ex- 
erted by the particles on each other are assumed to be in 
the lines joining the particles, and to be equal and opposite, 
according to Newton's laws. Hence in any summation of 
all the forces dT and f?iV, acting on all the particles, the 
reactions of the particles on each other would annul and 
leave only the summation of the impressed forces. There- 



268 



APPLIED MECHANICS FOR ENGINEERS 



fore the sum of the moments of the effective forces about 
the line AB is equal to the sum of the moments of the im- 
pressed forces about that line; i.e. 



/• 



rdT = 2(mom of impressed forces) = SPa. 

But dT= radM. 

.'. ^Pa = Cr'^adM= aCr'^dM, 



or 



'ZPa = « J, 



where / is the moment of inertia of the body about the 
axis of rotation. 

The name moment of inertia is suggested for I r^dM 

since it is seen to be equivalent to the moment of a force 
which would produce unit angular acceleration of the 

body about the given axis, 
against the inertia of the 
body. 

The work done by any 
force P during the rota- 
tion through the angle dd is 
Pcos(;)c?s(Fig. 215). But 

ds = rdd^ and cos (^ = - . 




Fig. 215 



and hence the expression 
for the work may be written j PadQ., and the total work 
done by all the impressed forces for any rotation becomes 

^(^Pa)de. 



WORK AND ENERGY 



269 



It was shown above that ^Pa = a/, and hence the work 
done may be written 



d 



CO 



1 1 adO, or since a= co 
^ du 

(Odo) = ^ X(w| — 0)2), 

where Wq and w^ are the initial and final angular velocities 
of the body for the given displacement. 

It follows that the work that the rotating body, with 
angular velocity coq, could do against resisting forces before 
coming to rest is I i^o- That is, the kinetic energy of a 
body rotating about a fixed axis with angular velocity a> is 
^ i&)2, where /is the moment of inertia of the body about 
the axis of rotation. 

As an illustration, let us consider 
the case of two weights (See Fig. 
216), a^=20 lb. and a^ = 10 lb., 
suspended from drums rigidly at- 
tached to each other and of radii 3 
ft. and 2 ft. respectively. Let the 
weight of the two drums and shaft 
be 644 lb., and the radius of gyration 
2 ft. The radius of the axle is one 
incli and the axle friction 30 lb. The 
friction acts tangentially to the axle. 

Assuming that the initial angular 
velocity &)q is one radian per second, and the final anguLar 
velocity 18 radians per second, how many revolutions will 
the drums make? 




Fig. 216 



270 



APPLIED MECHANICS FOR EJNGINEERS 



The external forces actiiiuf on the drains are the ten- 
sions, 7\ and T^^ in the cords attached to the weights G^ 
and G^ respectively, the reaction of tlie axle, and the force 
of friction. Acting on the weight G^ are the forces (tj 
and 2\ and on G2 the forces (rg and T^- The work-energy 
equations for these three bodies are respectively, 

2 7rr^7iT^ - 2 irr^^nT.^ - 2 Trr^n • 30 = ^ 80(18^ - 12), 
lirr^HiG^- 2\) 






2 7rr2<^2-^2) = ^^[a8r2)2-i], 
- 9 

where r^ is the radius of the large drum, r^ that of the 

small drum, and r^ that of the axle. 

Eliminating T^ and T^ by adding the three equations, 

2 irr-^nG^ — 2 Trr.^^i (^2 — 2 7^r37^ • 30 

= YsO +--1^ +-^Vl82 - 12). 
2V ^ ff J 

Substituting the known values, we obtain 7i = 59.5. 

Problem 353. In the above illustration, what are the velocities of 

G^ and G2 when w has its 
initial and final values ? In 
what time do the drums 
make the 59.5 revolutions? 

Problem 354. The drum 
in Fig. 217 is solid and has 
a radius 7* and a thickness h. 
Initially, it is rotating, mak- 
ing wo radians per second, 
but it is brought to rest by 
the action of a brake. The 
brake is applied from below by a force P acting at the end of the beam. 




WOBK AND ENERGY 271 

pi 

The force of friction between the drum and brake is — , where P' is 

4 

the normal pressure exerted by the beam on the drum. The radius 
of the axle is r^, and the axle friction (.05) P", where P" is the pres- 
sure of the axle on the bearing. Required the work -energy equation. 
Since the drum comes to rest, the final kinetic energy is zero, so 
that 

_ \l^^2^ + UL'2 Trrn + (.05) P"2 Tr/y = 0. 

There are no working forces, so we find the equation reducing to the 
form : the initial kinetic energy equals the work of resistance. The 
normal pressure exerted by the beam on the drum maybe found by 
taking moments about the hinge of the beam. Then 

p, ^ a +h p 
b 

The number of revolutions turned through in coming to rest is 
designated by n. The equation then becomes 

1 r 2 irrn (a -}- b) P , , n-\ nun 
-loii = ^^ — — h (-Oo) P"2 Trr,n- 

2 2 ^ ' V / 1 

Problem 355. Suppose the drum in the preceding problem to 
be 3 ft. in diameter, \\ in. thick, and made of cast iron. It is nuik- 
ing 4 revolutions per second when the force P = 100 lb. is applied to 
the beam. The length of the drum is 6 ft., and the rim weighs twice 
as much as the spokes and hub. If ^ = 1.25 ft., a = b = i ft., and 
Tj = 1 in., find the number of revolutions that the drum will make 
before coming to rest. Assume the friction of the brake on the drum 
to be I the normal pressure, and the friction of the axle (.05) P". 

Problem 356. The drnm in the preceding problem is making 
3 revolutions per second. What force will be required to bring it 
to rest in 100 revolutions? 

Problem 357. If the brake in Problem 355 is above instead of 
below the drum, how will the results in Problems 355 and 356 be 
changed? 



272 



APPLIED MECHANICS FOR ENGINEERS 



Problem 358. 



so as to rotate due to the weight G. 




^ 



(J= too LBS. 



FiCx. 218 



A square prism as shown in Fig. 218 is mounted 
The elastic cord runs over the 
pulley B and meets the square 
at P', but is connected to the 
square at P. The mechanism 
is such that motion begins when 
P is in the position tjhown, and 
ceases when tlie prism lias made 
a quarter turn ; that is, when P 
reaches P'. The diameter of 
the journal is 2 in., and the 
weight on the same is 600 lb. 
The force of friction on the journal is 60 lb., and on the pulley at B 
equivalent to 10 lb. acting at the rim of the pulley. Find the tension in 
the cord w^hen P reaches P'. The cord is elastic, and is made of such 
material that it elongates, due to a pull of 100 lb., .02 in. in each inch of 
length. What is the elongation per inch due to the fall of G as stated ? 

148. Brake-shoe Testing Machine. — The brake-shoe testing 
machine owned by the Master Car Builders' Association has 
been established at Purdue University. It consists of a 
heavy flywheel attached to the same axle as the car wheel. 
These are connected with the engine, and may be given any 
desired rotation. When this has been obtained, they may 
be disconnected and allowed to rotate. The dimensions and 
weight of the parts are known so that the kinetic energy of 
the flywheel and rotating parts may be computed by noting 
the angular velocity. When the desired velocity has been 
attained, the brake shoe is brought down on the car wheel. 
The required normal pressure on the shoe at A (See Fig. 
219) is obtained by applying suitable weiglits at B. The 
system of levers is such that one pound at B gives a nor- 
mal pressure of 24 lb. on the brake shoe. The weight of 



WORK AND ENERGY 



273 



the levers themselves gives a normal pressure of 1233 lb. 
Provision is also made for measuring the tangential pull of 
the brake friction ; this, however, is not shown in the fio-ure. 




f) 



Fig. 219 

The weight of the flywheel, car wheel, and shaft, and 
all rotating parts is 12,600 lb., and the radius of gyration 
is V2.16. The weight of 12,600 lb. is supposed to be the 
greatest weight that any bearing in passenger or freight 
service will be called upon to carry. The diameter of the 
flywheel is 48 in., its thickness 30 in., diameter of shaft 
7 in., and the diameter of the car wheel is 33 in. The 
brake-shoe friction is J the normal pressure of the brake 
shoe on the wheel, and the journal friction may be 
assumed as (.002) of the pressure of the axle on the 
bearing. The work-energy equation for the rotating 
parts after being disconnected from the engine becomes 



1 12,600 



33 



2'ikr*--^^'^'^'-"^'^ + (^-'^^-^-^^^i-^ii 



71 



+ (1233 + 12,600 H- 24 a)(.002)2 tt - 7i = 0. 

24 



274 



APPLIED MECHANICS FOR ENGINEERS 



Problem 359. The speed is such as to correspond to a speed of 
traiu of a mile a minute when brakes are applied. What must be 
the weight G so that a stop may be made in a thousand feet? What 
is the corresponding normal pressure on the brake shoe? 

Problem 360. If the speed corresponds to the speed of a train 
of 100 mi. per hour, what weight G would be necessary to reduce the 
speed to 60 mi. per hour in one mile? What is the normal pressure 
on the brake shoe necessary? 

Problem 361. If the velocity corresponds to a train velocity of 
60 mi. per hour, and the apparatus is brought to rest in 220 revolu- 
tions, the weight G is 100 lb. Find the tangential force of friction 
acting on the face of the w^heel. What relation does this bear to the 
normal brake-shoe pressure ? 

Note. In the preceding problems, the ratio (the coefRcieut of fric- 
tion, see Art. 135) has been taken as ^. One of the important uses 
of this testing machine is to determine the coefficient of friction for 
different types of brake shoes. Experiment shows that it varies 
generally from I to i, sometimes going as high as y\. 

149. Kinetic Energy of a Body having Plane Motion. — Sup- 
pose a body to have plane motion. Let the angular veloc- 
ity of the bod}' at a given 
instant be co and the ve- 
locity of the center of 
gravity of the body be Vy 
Choose coordinate axes 
so that the origin coin- 
cides at the given instant 
with the center of gravity 
and the a:-axis coincides 
with the direction of 
motion of tlie center of gravity, the x- and z-axes being in 
the plane of motion of the body (Fig. 220). 




Fig. 220 



WORK AND ENERGY 275 

Let dM be any element of mass of the body, distant r 
from the ^-axis. The velocity of dM is then composed of 
the velocity of any point on tlie ^-axis and the velocity of 
c?ili" relative to the ^-axis. Hence, v, the velocity of dM, 
is given by 

v^ = v'^-{- r^tiP' -|- 2 v-^ro) cos <^, 

where ^ is the angle whicli r makes with a line parallel to 
tlie ;3-axis. But r cos </> = 2, so that 

v^ — v\ 4- r^ft)^ + 2 v^(tiz. 

The kinetic energy of the whole body is the sum of the 
kinetic energy of its particles, or 

K. E. of Body = ^\{v\ + r'^o? -h 2 v^uiz)dM 

= I v\CdM+ I ay'-jrUM^ 2 v^coCzdM. 

Since the a;?/-plane passes through the center of gravity, 

\dM= 0. (Art. 33.) 



/' 



Therefore 

K. E. of Body = \ Mvi + 1 Iu>^ 

where M is the mass of the body and / its moment of 
inertia about a gravity axis perpendicular to the plane of 
motion. 

This formula may be expressed in words as follows : 
The kinetic energy of a body having plane motion is equal 
to the Icinetie energy the tvhole mass would have if concen- 
trated at the center of gravity, ivith the velocity of the center 
of gravity, plus the kinetic energy of rotation that the body 



276 



APPLIED MECHANICS FOR ENGINEERS 



would have if the gravity axis were at rest and the body 
rotating about it ivith the same angular velocity. 



150. Work and Kinetic Energy in Plane Motion. — If im- 
pressed forces act on a body liaviiig plane motion, the 
work done by tliese forces in any displacement equals the 
change in kinetic energy of the body in that dis- 
placement. 

Proof: The total gain in the kinetic energy of the body 
is the sun:i of the increments in kinetic energy of all the 
particles of the body, which, by Art. 143, is the total 
work done on the particles by all the forces acting on 
them. In finding the total work done on the particles 
the work done by the forces that the particles exert on 
each other adds up to zero and there is left only the work 
done by the impressed forces. For two particles of the 

body, m-^ and TTig, remain the same 
distance apart and exert on each 
other forces that are equal and 
opposite and in the line joining 
the particles. During any dis- 
placement m^ has a motion made 
up of the motion of m^ and a 
motion about m-^ as a center (Fig. 
221). For that component of its motion parallel to the 
motion of m^ the forces exerted by the particles on each 
other are equal and opposite and have the same displace- 
ment, and hence the work done by one force is just equal 
to the work done against the other. For tlie motion of 
rotation of m^ about m^ the force acting on m^ is perpen- 




FiG. 221 



WORE AND ENERGY 



211 



dicular to the direction of displacement, and hence no work 
is done. Hence the forces exerted by the particles on each 
other do no work, and we may write for any plane motion. 

Work doue by impressed forces = change in kinetic energy of body. 

As an illustration, consider a body of circular section, 
as a hoop, cylinder, or sphere, with center of gravity at 
the center of the circular section, rolling without slipping 
down an inclined plane (Fig. 222). The impressed 
forces acting on the body are 
its weight, 6r, the normal 
reaction of the plane, iV, and 
a retarding friction force, i^, 
along the plane. The point 
of application of iV has no 
displacement in tlie direction 
in which JV acts, and hence no work is done by N. The 
point of application of F^ continually changing in the body, 
has at each instant a motion at right angles to F; for if 
there is no slipping, the point of the body in contact with 
the plane at any instant leaves the plane at right angles to 
the plane. Hence no work is done by F. The total work 
done in the descent is therefore Gh. 

If (Oq and Vq are the angular velocity of the body and 
the linear velocity of its center of gravity respective!}^ at 
the top of the plane and co and v the corresponding values 
at the foot, then the work-energy equation is 

I 70)2 + .] Mv^ - 1 IcD-^ -IMvl^ ah. 




Fig. 222 



If 8 is the distance passed through by the center and 6 



278 APPLIED MECHANICS FOR ENGINEERS 

tlie angle turned through in the same time, 

s = rO. 

Therefore 

or 

Substituting 

the work-energy equation becomes 







ds 
dt 


d6 
"dt' 












V = 


ro). 








ft) 


_ ^ 

— ~ -> 


(Oq = 


\i= 


^k\ 


M= 


a 




r 




r 


y 




.9 



2gh 



or ly^ — v^ = . 

Problem 362. Prove that all .solid spheres will roll down the 
inclined plane at the same rate. Find the velocity at the foot of the 
plane. 

Problem 363. A uniform sphere, a uniform disk, and a hoop, 
starting at the top of an inclined plane, roll from rest to the foot. 
Find the velocity of each on reaching the foot. In what order do 
they arrive? 

Problem 364. Which will roll faster down an inclined plane, a 
hollow sphere with diaineter of the hollow one half that of the 
sphere, or a solid uniform disk? 

151. Kinetic Energy of Rolling Bodies. — It is convenient 
to express the kinetic energy or combined rotation and 
translation of such bodies as rolling wheels in a different 
form from tliat given in the preceding article. There is 
some mass M^ that will have the same kinetic energy 
when translated with a velocity v^ as tlie kinetic energy 



WOUK AXIJ EX ERG Y 279 

of translation plus the kinetic energy of rotation of tlic 
body of mass 31; that is, 

2 2 2' 

For a wheel rolling on a straight track cor = t^j, where r is 
the radius. 

Then 3L=M-h{. 

This has been called tlie equivalent mass. 

For example, for a rolling disk, since 1= ^ Mr^, 31^ = | 31, 

Problem 365. A sphere rolls without slipping down an inclined 
plane. Show that its kinetic energy is the same at any instant as that 
of a sphere whose mass is f larger translated with a velocity equal to 
the velocity of the center of gravity of the rolling sphere. 

Problem 366. The sphere in the preceding prohlem is made of 
steel, 12 in. in diameter, and the inclination of the plane is 30°. If 
Vq = 10 ft. per second, what will be the velocity 10 ft. down the plane? 

152. "Work-energy Relation for Any Motion. — The rela- 
tion between work and energy for the motions considered 
in this chapter holds for more complicated motions and 
for motions in general. The limits of the present work 
will not admit the proof of the general theorem. It may 
be said, however, that for any motion the work done by 
the working forces equals the work done by the resisting 
forces plus the change in kinetic energy. In the case of 
the motion of a complicated machine, the work done by 
the working forces equals the work done against the resist- 
ances plus the gain in kinetic energy of the various parts 
of the machine. 



280 



APPLIED MECHANICS FOE ENGINEERS 



153. Work Done when Motion is Uniform. — When the 
motion is uniform, the change in kinetic energy is zero, 
and tlie work-energy equation reduces to the form : work 
done equals the work done against the resistance overcome. 

As an illustration, let us consider the case of a loco- 
motive moving at uniform speed and represented in Fig. 
223. Suppose P the mean effective steam pressure (See 
Art. 139), F the friction of the piston, F' the friction of 
the crosshead, F" the journal friction, F'" the crank-pin 




nizn 



Fig. 223 

friction, T the friction on the rail, M the draw-bar resist- 
ance, M' the horizontal component of pressure of the 
drivers' axle on the frame, r, r-^, and r^ the radii of the 
crank-pin circle, the driver axle, and the crank-pin re- 
spectively, 6r the weight of the locomotive, and N' and N 
the normal reactions of the rails on the wheels. Consider 
both sides cf the locomotive and write the work-energy 
equation for a distance s, equal to a half turn of the driver 
(from dead center A to dead center J5), that the locomo- 
tive travels. 

Considering the work done on the frame, counting both 
cylinders, we have 

E'lra -f- 2(i^-f F')iTa -\- 2 F"7ra = (2 P + R)7ra. (1) 



WORK AND ENERGY 281 

Considering the work done on the rotating and oscillat- 
ing parts, 

4- 2F"(nTa+7Tr^}^2F"''Trr^. (2) 

Adding these equations there results 

4jPr = R^ra +{F+F')^r + 2F"TTri + 2 /^'"-rrr^. (3) 

If we neglect friction, this equation becomes 

4 Pr = iraR^ 
or r = ^^ R. 

Here P is the mean effective pressure in one cylinder. 

This is the formula usually given for the tractive power 
of a locomotive having single expansion engines. This 
may be expressed in terms of the dimensions of the cylin- 
ders and the unit steam pressure. Let p be the unit 
steam pressure in pounds per square inch, I the length of 
the cylinder in inches, d the diameter of the cylinder in 
inches, and d-^ the diameters of the drivers in inches ; then 



R = 



d?pl 



For uniform motion the train resistance cannot exceed 
the friction force or force of adhesion between the drivers 
and the rails, since these are the external horizontal forces 
acting on the engine at any time. This force of adhesion 
in American practice is usually taken as \ or -J of the 
weight on the drivers. 

Problem 367. Derive equation (3) of this article by considering 
the work done on the whole engine. 



282 APPLIED MECHANICS FOR ENGINEERS 

Problem 368. What resistance R may be overcome by a locomo- 
tive mo\ iiig at uniform speed, diameter of drivers 62 in., cylinders 
16 X 24 in., and a steam pressure on the piston of 160 lb. per square 
inch ? What should be the weight of the locomotive on the drivers? 

Problem 369. If the diameter of the drivers of a locomotive is 
68 in., and the size of the cylinder is 20 x 24 in., what train resistam^e 
may be overcome by a steam pressure of 160 lb. per square inch? 

Problem 370. A locomotive has a weight of 155 tons on the drivers. 
If the adhesion is taken as |, this allows 31 tons for the drawbar pull. 
The train resistance per ton of 2000 lb., for a speed of 60 mi. per hour, 
is 20 lb. Find the weight of the train that can be pulled by the 
locomotive at the speed of 60 mi. per hour. 

Problem 371. An 80-car freight train is to be pulled by a single 
expansion locomotive at the rate of 30 mi. per hour. The weight of 
each car is 60,000 lb., and the resistance for this speed is 10 lb. per 
ton. "What must be the weight on the drivers, if the adhesion is ^? 



CHAPTER XIII 

FRICTION 

154. Friction. — When, one body is made to slide over 
another, there is considerable resistance offered because of 
the roughness of the two bodies. A book drawn across 
the top of a table is resisted by the roughness of the two 
bodies. The rough pai'ts of the book sink into the rough 
parts of the table so that when one of the bodies tends to 
move over the other, the projections interfere and tend to 
stop the motion. The bearings of machines are made 
very smooth, and usually we do not think of such surfaces 
as having projections. Nevertheless they are not perfectly 
smooth, and when one surface is rubbed over the other, re- 
sistance must be overcome. This resisting force to the 
motion of one body over another is known as friction. 
When the bodies are at rest relative to each other, the 
friction is known as the friction of rest, or static friction. 
When they are in motion with respect to each other, the 
friction is known as t\\Q friction of motion, ov kinetic friction . 

155. Coefficient of Friction. — If the body represented in 
Fig. 224 be pulled along the horizontal plane by the force 
P, the following forces will be acting on it : the downward 
force Gr and the reaction R inclined back of the vertical 
through the angle 6. Th<i reaction R of the plane on the 
body may be resolved into two components, one horizontal 

283 



284 



APFLIED MECHANICS FOR ENGINEERS 



and one vertical. The horizontal force is known as the 
force of friction, and the normal force, the normal pressure. 

The tangent of the angle 




e, or 1^, is 



jailed the 



coefficient of friction. This 
coefficient of friction, 
which we shall represent 
by /, may be defined as 
the ratio of the force of 

friction to the normal pressure ; it is an absolute number. 
The angle 0, between R and the normal to the surface 

of contact, is called the angle of friction. 

(9=tan-i/. 

The coefficient of friction is usually determined by al- 
lowing a body to slide down an inclined plane, as shown in 
Fig. 225. The angle 6 is in- 
creased until the force of friction 
F will just keep the body from 
sliding down the plane. The 
angle 6 is then called the aiigle of 
repose., and the tangent of 6 is the 
coefficient of friction. 

Proof: When the body is just 
on the point of slipping down, 
the force R must just balance G. Hence the angle be- 
tween R and N is equal to the angle of inclination of 
the plane, or the angle of friction is equal to the angle of 
inclination of tlie plane. 




FRICTION 285 

It is possible with such an apparatus to determine the 
coefficient of friction for various materials. It has been 
found that after motion begins tlie friction is less; that is, 
the friction of motion is less than the friction of rest. This 
is an important law for engineers. 

156. Laws of Friction for Dry Surfaces. — Very little was 
known of the laws of friction until within the last seventy- 
live years. About 1820 experiments were made that 
seemed to show that, for such materials as wood, metals, etc., 
friction varies with the pressure, and is independent of 
the extent of the rubbing surfaces, the time of contact, 
and the velocity. A little later (1831) ]\Iorin published 
the following three laws as a result of his experiments on 
friction : 

(1) The friction between two bodies is directly propor- 
tional to the pressure ; that is, the coefficient of friction is 
constant for all pressures. 

(2) The coefficient and amount of friction for any given 
pressure is independent of the area of contact. 

(3) The coefficient of friction is independent of the ve- 
locity., although static friction is greater than kinetic fric- 
tion. 

These laws of Morin hold approximately for dry, unlu- 
bricated surfaces, although it has been found that an in- 
crease in speed lowers the coefficient of friction. The 
coefficient of friction is a little greater for light pres- 
sures upon large areas than for great pressures on small 
areas. 

The following is a table of some of the coefficients of 
friction as determined by Morin : 



'2Si.\ 



AV PLIED MECHANICS FOR EXCINEEIiS 



Coefficients of Frk tiox. dtk to ^roiuN 



Material 


Condition of Surfack 


£ 1 


5 






M 2 
O fc PS 


hJ 


Brick on limestone 


Dry 


.67 


33^ 50' 


Cast irou on cast iron 


Slightly greased 


.16 


9° 6' 


Cast iron on oak 


Wet 


.65 


33'' 2' 


Copper on oak 




.17 


9^38' 


Copper on oak 


Greased 


.11 


6° 17' 


Leather on cast iron 




.28 


15° 39' 


Leather on cast iron 


Wet 


.38 


20^ 49' 


Leather on cast iron 


Oiled 


.12 


•6° 51' 


Leather on oak 


Fibers parallel 


.74 


36° 30' 


Leather on oak 


Fibers crossed 


.47 


25° 11' 


Oak on oak 


Fibers parallel, dry 


.62 


31° 48' 


Oak on oak 


Fibers crossed, dry 


.54 


28° 22' 


Oak on oak 


Fibers parallel, soaped 


.44 


23° 45' 


Oak on oak 


Fibers crossed, wet 


.71 


35° 23' 


Oak on oak 


Fibers end to side, dry 


.43 


23° 16' 


Oak on oak 


Fibers parallel, greased 


.07 


4° 6' 


Oak on oak 


Heavily loaded, greased 


.15 


8M5' 


Oak on pine 


Fibers parallel 


.67 


33° 50' 


Oak on limestotie 


Fibers on end 


.63 


32° 15' 


Oak on hemp cord 


Fibers parallel 


.80 


38° 40' 


Pine on pine 


Fibers parallel 


.56 


29° 15' 


Pine on oak 


Fibers parallel 


.53 


27° 56' 


Wrought iron on oak 


Wet 


.62 


31° 48' 


Wrought iron on oak 




.65 


33° 2' 


Wrought iron on wrought iron 




.28 


15° .39' 


Wrought iron on cast iron 




.19 


10'^ 46' 


Wrought iron on limestone 




.49 


26° 7' 


AVood on metal 


Greased 


.10 


6° 0' 


Wood on smooth stone 


Dry 


.58 


30° 7' 


Wood on smooth earth 


Dry 


.33 


18^ 16' 



FRICTION 



287 



Problem 372. Find the force ]^ necessary to move with uniform 
velocity a weight of 100 lb. up a plane inclined 30^ to the horizontal 
(«) when P is horizontal, (b) when parallel to the plane, (c) when in- 
clined at an angle of 60° to the horizontal, given that the coefficient of 
friction between the weight and the plane is .20. Find the force just 
necessary to prevent the body from sliding down the plane in each of 
the cases. 

R 





Fig. 226 



Problem 373. Show that the force P, inclined at an angle to 
the plane, that will {a) just move the weight up the plane, (/>) just 
prevent it from sliding down the plane (Fig. 226), is 

sin(a + ^) sin(oc- ^ ) 

(d) r = 7-r- /^ (r, (n) F = 7-7 /. (jr, 

^ ^ COS(cf> - 6) ^ ^ COS ((f>+ 6) 

where 6 is the antiie of friction. 



Problem 374. Show that the least values of P in the preceding 
problem are when <^ = ^ in (a), and when ^ = — ^ in (//) ; i.e. when 

(a) P = G' sin (a + 0), (h) P = G sin {a - 0). 

Problem 375. In Fig. 227 the weight 
G is raised l)y the horizontal force P. If 
the only friction is between the surfaces 
of the wedge and the weight, prove that 
the value of P just sufficifut to raise the 
weight is 




Fio. 227 



P =G tiiu((i + 6). 



288 



APPLIED MECHANICS FOB ENGINEERS 



Problem 376. Defining the efficiency, E, of the wedge as the ratio 
of the useful work accomplished in raising the weight to the total 
work done by P (Fig. 227) show that 

P _ /tan a 
tan (a -\- 0) 

Problem 377. For a given value of 0, show that E isa maximum 
when 

a = 450 _ ^ . 
2 

(Since a square-threaded screw may be regarded as an inclined plane, 

this formula also holds for such a 

screw.) 

Problem 378. Find the value 
of the horizontal force P that will 
just raise the weight of 500 lb. in 
Fig. 228, given that the coefficients 
of friction at J., J5, and C are re- 
spectively .20, .25, and .30. 




■//////^//////yy//. 



Fig. 228 



Suggestion. Consider the wedge 
and block separately, and the forces that hold them in equilib- 
rium. 

157. Friction of Lubricated Surfaces. — The laws of fric- 
tion, as given by Morin and stated in the preceding article, 
hold approximately for rubbing surfaces, when the sur- 
faces are dry or nearly so ; that is, for poorly lubricated 
surfaces. If, however, the surfaces are well lubricated so 
tliat the projections of one do not fit into the other, but 
are kept apart by a film or layer of the lubricant, the laws 
of Morin are not even approximately true. The study of 
the friction of lubricated surfaces, then, may be divided 
into two parts : (1) tlie study of poorly lubricated bear- 
ings, and (2) the study of well lubricated bearings, the 



FRICTION 289 

friction of which varies from J to ^ that of dry or poorly 
lubricated bearings. 

Since the friction of poorly lubricated bearings is about 
the same as that of dry surfaces, we shall consider that 
the laws of Morin hold, and shall confine our attention to 
the friction of well lubricated bearings. If the lubricant 
is an oil, the friction of the bearing is no longer due to 
one surface rubbing over the other, but to the friction 
between the bearing and the oil, and to the internal fric- 
tion of the oil. That is, the oil adheres to the two sur- 
faces, and its own particles attract each other, and the 
motion of one of the surfaces with respect to the other 
clianges the positions of the oil particles. It is to be 
expected, then, that the friction of an oiled bearing will 
depend upon the viscosity of the oil^ upon the thickness of 
the layer interposed between the surfaces^ and upon the 
velocity and form of the bearing. 

The coefficient of friction is no longer constant, but 
varies with the temperature, velocity, and pressure. The 
variation of the coefficient of friction of a paraffine oil 
with temperature is shown in Fig. 229 when the pressure on 
the bearing is 33 lb. per square inch and a velocity of rub- 
])ing of 296 ft. per minute. It is seen that the coefficient 
of friction decreases with increase of temperature until a 
temperature of 80° F. is reached, when it increases rapidly. 
This means that above this temperature the oil is so thin 
that it is squeezed out of the bearing, and the conditions 
of dry bearing are approached. Tlie temperature at 
which oils show an increasing coefficient of friction is dif- 
ferent for different oils, even at the same pressure and 



290 



APPLIED MECHANICS FOR ENGINEERS 



velocity. The curve in Fig. 229, however, may be re- 
garded as typical of all oils when the pressure and velocity 
are constant. 



w 



&0 



60; 



50 



40 



30 





— 


: 

-: — r 


— 


~ 


- 








' 




















— 


— 


— 


— 







■- ■ 


— 


I^ 





Ip 


I^ 


^ 




— 


-t^- 


































^1 










_:_- 


„_. 


__„. 







, 




_: 


. 





1 









if^ 


^— 


T- 


— 


, 


— 


— 


— 


— 


^— 


\ 


— " 


— 


- 


— 




" i - 




-re 


— 


— 


— 


— - 


— 




— 


— 


" 


— 


" 


!___ 


— 


— 


:. 1 




■4 










i 
1 


— 


— r 




-- 






1 
1 


— 


. 


1 -;- ""■ ;3t 


zzv. 






-^ 


" : 1 




\ 




w 






^~ 


■ r 








~" 


._ 


- 


~ 








1 — h— 


TTT' 




















_ 


;\^ 


_ 


























- , 




V 










f:3 


















1 










\ 








^i 


i?? 


cc 
p 

V 


DEF 


RES 
ELO 


FIC 
F A 

3UR 
CIT 


! 




)N 

H 

E 






\: 




::± 


par; 

E 33 LB 
r 296 F£ 

- . 1 . 


ui- 
\FF 

S. F 

:et 


1- 

INf 

ER 
PEF 


RIC"" 
-. 

SQ. 
Ml 


1 K 

IL 

INC 

SUT 


■~~ 


— ■ 


— 


s 


K 




I 


— 


E^ 


|:r:^ 




1^ 


' 


— 


— 


— 


— 


^ 


— 


Six; 


\^ 


\^ 


t£. 


riilrifr 


\^ 





.01 .02 .03 

coefficient of friction 

Fig. 229 



.01 



The following table, due to Professor Thurston, shows 
the relation between the coefficient of friction and tem- 
perature for a sperm oil in steel bearings when the veloc- 
ity of rubbing is 30 ft. per minute: 



FRICTION 



291 



PRESStTRE, Lb. 


Temperature, 


Coefficient 


Pressure, Lb. 


Temperature, 


Coefficient 


I'ER Sq. In. 


Degrees F. 


OF Friction 


PER Sq. In. 


Degrees F. 


OF Friction 


200 


150 


.0500 


100 


110 


.0025 


•200 


140 


.0250 


50 


110 


.0035 


200 


130 


.0160 


4 


110 


.0500 


200 


120 


.0110 


200 


90 


.0040 


200 


110 


.0100 


150 


90 


.0025 


200 


100 


.0075 


100 


90 


.0025 


200 


95 


.0060 


50 


90 


.0035 


200 


90 


.0056 


4 


90 


.0400 


150 


110 


.0035 









It is seen that for a pressure of 200 lb. per square inch 
as the temperature increases from 90°' F. the coefficient in- 
creases, indicating that the temperature of 90°, for the 
given pressure and velocity, was above the temperature 
at which the oil became so thin as to be squeezed out and 
the bearing to approach the condition of a dry bearing. 
For a constant temperature 110° F. and 90° F. the coeffi- 
cient is seen to decrease with increase of pressure up to a 
certain point and then to increase. This is a typical be- 
havior of oils when tlie temperature is constant and the 
pressure varies. 

At speeds exceeding 100 ft. per minute, the same author- 
ity found '' that the heating of the bearings within the 
above range of temperatures decreases the resistance due 
to friction, rapidly at first and then slowly, and gradually 
a temperature is reached at which increase takes place and 
progresses at a rapidly accelerating rate." 

The relation between the coefficients of rest and of motion 
as determined by Professor Thurston for three oils is given 



292 



APPLIED MECHANICS FOR ENGINEERS 



below. The journals were cast iron, in steel boxes ; velocity 
of rubbing 150 ft. per minute and a temperature 115° F. 





Sperm Oil 


West Virgini 


A Oil 


Lard 


Pressire, 


At 150 


At 


At 


At 150 


At 


At 


At 150 


At 


At 


Lb. per 


ft. per 


start- 


stop- 


ft. per 


start- 


stop- 


ft. per 


start- 


stop- 


Sy. In. 


min. 


ing 


ping 


min. 


ing 


ping 


mm. 


ing 


ping 


50 


.013 


.07 


.03 


.0213 


.11 


.025 


.02 


.07 


.01 


100 


.008 


.135 


.025 


.015 


.135 


.025 


.0137 


.11 


.0225 


250 


.005 


.14 


.04 


.009 


.14 


.026 


.0085 


.11 


.010 


500 


.004 


.15 


.03 


.00515 


.15 


.018 


.00525 


.10 


.016 


750 


.0043 


.185 


.03 


.005 


.185 


.0147 


.0066 


.12 


.020 


1000 


.009 


.18 


.03 


.010 


.18 


.017 


.0125 


.12 


.019 



Steel Journals and Brass Boxes 



500 
1000 



.0025 
.008 












.004 
.009 





It is seen that the coefficient of friction at starting is much 
greater than at stopping, and that these are both much 
greater than the value at a speed of 150 ft. per minute. 

For an intermittent feed such as is given by one oil hole, 
without a cup, oiled occasionally. Professor Thurston found 
for steel shaft in bronze bearings, with a speed of rubbing 
of 720 ft. per minute, the following coefficients of friction : 



Oil 


Pkessure, Lb. per Sy. In. 




8 


16 


32 


48 


Sperm and lard .... 
Olive and cotton seed 
Mineral oils 


.ir)9-.25 

.l(J0-.283 

.154-.261 


.138-. 192 
.107-.245 
.145-.233 


.086-.141 
.101-.168 
.086-.178 


.077-.144 
.079-.131 
.094-.222 









FRICTION 293 

The results show that the coefficient decreases with the 
pressure within the range reported, but that the results 
are considerably higher than those for well lubricated 
bearings. He also found in connection with the same 
tests that with continuous lubrication sperm oil gave the 
following coefficients : 

Pressure, Coefficient 

Lb. per Sq. In. of Friction 

50 .0034 

200 .0051 

300 .0057 

The results of tests of the friction of well-lubricated 
bearings are summarized by Goodman (^Engineering Neivs^ 
April 7 and 14, 1888) as follows : 

(a) The coefficient of friction of well lubricated surfaces is 
from ^ to Jq- that of dry or poorly lubricated surfaces. 

(5) The coefficient of friction for moderate pressures and 
speeds varies approximately inversely as the normal pressure ; 
the frictional resistance varies as the area in contact^ the nor- 
mal pressure remaining the same. 

(c) For low speeds the coefficient of friction is abnormally 
high., but as the speed of rubbing increases from about 10 to 
li)0 ft. per minute, the coefficient of friction diminishes, and 
again rises when that speed is exceeded, varying approxi- 
mately as the square root of the speed. 

(f?) The coefficient of friction varies approximately in- 
versely as the temperature, within certai?i limits; namely, 
just before abrasion takes place. 

158. Method of Testing Lubricants. — To make the matter 
of the tests of the friction of lubricants clear, it will be 



294 



APPLIED MECHANICS FOR ENGINEERS 



convenient to make use of the description of a testing 
machine used by Dean W. F. M. Goss at Purdue Uni- 
versity on graphite, and a mixture of graphite and sperm 




Fig. 230 

oil. In making the tests the apparatus shown in Figs. 230 
and 231 was used. (See " A Study in Graphite,'' Joseph 
Dixon Crucible Co.) 

This apparatus represents, in principle, the machines 
generally used for testing lubricants. It is therefore 
shown in some detail. The weight Gr is hung from the 
shaft upon which it is suspended by the form of box to be 
tested. The desired speed of rubbing is obtained by 
means of the cone of pulleys, and the pressure on the bear- 
ing is adjusted by the spring. The temperature of the 
bearing is read from the thermometer inserted in the bear- 
ing. When rotation takes place, the weight Gr is rotated 
a certain distance dependent upon the friction. This dis- 
tance is measured on tlie scale A. The forces acting upon 



FRICTION 



295 



R-\-0 



the peiiduluin G- are shown in Fig. 2ol, where R represents 
the resistance of the spring, F tlie force of friction, I the 
distance of the center of gravity of G 
from the axis of rotation, (^ the angle 
through which G is deflected, r the 
radius of the shaft, and / the coeffi- 
cient of friction. Taking moments 
about the center of the shaft, we have, 
when G is lield in the position shown, 
due to the friction, 



or 



r(F^ F^)=Gl^m(i>, 
rf{R + G + B)=Gl sin (/>. 



/ = 



Gl sin ^ 

7^(2 R^G) 




Fig. 231 



It is customary to take G small 
compared with R, so that the pressure 
on both sides of the bearing may be considered equal to 
R^ tlie resistance of the spring. The formula then becomes 

f _ Gls'm <f> 

The spring is easily calibrated so that R may be made any- 
thing desired by compressing the spring through the ap- 
propriate distance, as indicated on the scale J^(Fig. 230). 
The quantities G, I, r, and R are known, and </> can be 
re-ad so that /can be calculated. 

The results of tests made upon a mixture of graphite 
and oil as a lubricant are given in the pamphlet. The 
tests were run under 200 lb. per square inch pressure, at 
a speed of rubbing of 145 ft. per minute. Oil was dropped 



296 



APPLIED MECHANICS FOR Ei\G INFERS 



into the bearing at the rate of about 12 drops per minute, 
showing a coefficient of friction of |. 

Problem 379. If the weight of the pendulum is 360 lb., the 
diameter of the shaft -i^ in., distance of the center of gravity of G 
from the center of shaft 2 ft., the angle (f> 5 degrees, and the average 
resistance of the spring 1000 lb., find the coefficient of friction. The 
weight G should not be neglected in this case. 

159. Rolling Friction. — The resistance offered to tlie 
rolling of one body over another is known as rolling fric- 
tion. It is entirely different from sliding friction, and its 

laws are not so well 
understood. When a 
wheel or cylinder (Fig. 
232) rolls over a track, 
the track is depressed 
and the wheel dis- 
torted. The force P 
necessary to overcome this depression and distortion is 
known as rolling friction. 

The forces acting on the wheel are seen from Fig. 232 
to be: P the working force, TTthe weight on the wheel, 
and B the reaction of the track or roadway. This upward 
pressure E is not quite vertical, but has its point of ap- 
plication a short distance K' from the vertical. Its line 
of action passes through the center of the wheel. The 
distance K' depends chiefly upon the roadway ; it is called 
the coefficient of rolling friction. It is measured in inches 
and is not a coefficient of friction in the strict sense that/ 
is the coefficient of sliding friction. 

Taking moments about the point of application of i2, 




Fig. 232 



FRICTION 



297 



we have, approximately, 



so that 



K' = 



WK' = Pr, 
Pr ^,. „ K'W 



W 



or r 



r 




When the track or roadway is ehistic or nearly so, we 
have a condition something like that represented in Fig. 
233. The wheel 
sinks into the ma- 
terial and pushes it 
ahead, at the same 
time it comes up 
behind the wheel. 
For a portion of 
the wheel on each 
side of the point 
the roadway is simply compressed ; over the remainder of 
the surface in contact, however, slipping occurs, as indi- 
cated by the arrows. The resultant resistance, however, 
is in front of the vertical through the center, and we have, 
as in the case of imperfectly elastic roadways, 

p_K'W 
r 

It lias been found by Reynolds (See Phil. Trans. Royal 
Soc, Vol. 166, Part 1) that when a cast-iron roller rolls 
on a rubber track, the slippage, due to the elasticity of the 
track, may amount to as much as .84 in. in 34 in. An 
elastic roller rolling on a hard track will roll less than the 
geometrical distance traveled by a point on the circumfer- 
ence. When the roller and track are of the same material, 
the roller rolls through less than its geometrical distance. 



298 



APPLIED MECIIAMCS FOR ENGINEERS 



160. Antifriction Wheels. — The axle A, of radius r, 
carrying a weight W, rests upon two wheels of radius r^^ 

turning on axles of radius r^ (Fig. 
234). 

The force on each of the bear- 
ings of wheels B and is 

W 

2cosy8' 

and if F is the friction at each of 

the bearings of B and C, and / 

Fig. 234 the coefficient of sliding friction, 

2cOSy8' 

and the work lost in friction at the bearings of B and (7 in 
one revolution of the axle A is 




/■ 



W 



z irr 



cos/3 ^ r^ 

Since the axle A rolls on B and (7, there is no sliding fric- 
tion there and the rolling friction is small enough to be 
neglected. 

If A were in an ordinary bearing, the work lost per 

revolution would be 

flirrW. 

Therefore the ratio of work lost with antifriction wheels 
to the work lost with plain bearing is 

r^ cos yS 
This ratio decreases as the ratio r^r^ increases and as /8 
decreases. 



FRICTION 299 

Problem 380. If W — 4 tons, the radius of the shaft is 2 iu., 
and the coefficient of friction is .07, what work is lost per revohition? 
If the shaft makes 3 revolutious per second, w^hat horse power is lost 
in friction ? Given also (3 = 45°, )\ = f in., and r^ = 4 in. 

Problem 381. In the case of the shaft mentioned in the preced- 
ing problem, how much more horse power would it take if the hear- 
ing were plain? What value of /3 would give the same loss due to 
friction in both the plain bearing and the one provided with friction 
wheels ? 

161. Resistance of Ordinary Roads. — Resistance to trac- 
tion consists of axle friction, rolling friction, and grade 
resistance. Axle friction varies from .012 to .02 of the 
load, for good lubrication, according to Baker. The 
tractive power necessary to overcome axle friction for 
ordinary American carriages has been found to be from 
3 lb. to 3| lb. per ton, and for wagons with medium-sized 
wheels and axles from 3| lb. to 4^ lb. per ton. 

The total tractive force per ton of load, for wheels 50 in., 
30 in., and 26 in., in diameter, respectively, is, according 
to Baker (^Engineering News^ March 6, 1902) : 



Tractive Force 
IN Pounds 



On macadam roads 

On timothy and blue grass sod, dry, grass cut . 
On timothy and V)lue grass sod, wet and springy 
On plowed ground, not harrowed, dry and cloddy 



57 
132 
173 
252 



61 
145 
203 
303 



70 
179 
288 
374 



Rolling resistance is influenced by tlie width of the tire. 
According to Baker, poor macadam, i)oor gravel, compres- 
sible earth roads, and, on agricultural lands, narrow tires, 



300 



APPLIED MECHANICS FOR ENGINEERS 



usually require less traction. On earth roads composed of 
dry loam with 2 to 3 in. of loose dust, traction with li-in. 
tires was 90 lb. per ton, and with 6-in. tires 106 lb. per 
ton. On the same road when it was hard and dry, with no 
dust, that is, when it was compressible, the traction was 
found to be 149 lb. per ton with l|-in. tires and 109 lb. 
per ton with 6-in. tires. On broken stone roads, hard and 
smooth, with no dust or loose stones, the traction per ton 
was 121 lb. witli IJ-in. tires, and 98 lb. with 6-in. tires. 
Moisture on the surface or mud increases the traction. 

Morin found that with 44-in. front and 54-in. rear 
wheels on hard dry roads the traction per ton was 114 lb. 
with either l|-in. or 3-in. tires. On wood-block pave- 
ments the traction per ton was 28 lb. with l2-in. tires, 
and 38 lb. with 6-in. tires. 

On asphalt, bricks, granite, macadam, and steel-road 
surfaces, investigated by Baker, the traction per ton 
varied from 17 lb. to 70 lb., the average being 38 lb. 

Morin gives the coefficient of rolling friction for wagons 
on soft soil as .065 in., and on hard roads .02 in. Accord- 
ing to Kent (" Pocket-Book "), tests made upon a loaded 
omnibus gave the following results : 



Pavemknt 



Granite 

Asphalt 

Wood 

Macadam, graveled 
Macadam, granite, new 



Speed, Miles 
PER Hour 



2.87 
3.56 
a.34 
3.45 
3.51 



Coefficient, 
Inches 



.007 

.0121 

.0185 

.0199 

.0451 



Resistance, pkb 
Ton, in Lb. 



17.41 
27.14 
41.60 
44.48 
101.09 



FRICTION 



301 



Problem 382. Compare the resistance oifered to a load of two 
tons pulled over asphalt, macadam, good earth roads, or wood-block 
pavement. ^Vidth of tires, fi in. 

162. Roller Bearings. — In the roller bearings the shaft 
rolls on hardened steel rollers as shown in cross section in 
Fig. 235. The roll- 
ers are kept in place 
in some way similar 
to that shown in the 
journal of Fig. 236. 
Such bearings are 
used where heavy 
loads are to be car- 
ried. Tests of roller 
bearings have been made by Dean C. H. Benjamin 
(^Machinery^ October, 1905), who determined the follow- 
ing values for the coefficient of friction. Speed 480 
revolutions per minute. 




DlAMF.TER OK 

JouRXAi-, IN Inches 


Roller Bearing 


Plain Cast-iron Bearing 


Max. 


Mill. 


Average 


Max. 


Min. 


A.verafre 


IB 


.036 


.010 


.026 


.160 


.099 


.117 


2A 


.052 


.034 


.040 


.129 


.071 


.094 


^^ 


.041 


.025 


.030 


.143 


.076 


.104 


2H 


.053 


.049 


.051 


.138 


.091 


.104 



It was found that the coefficient of friction of roller 
bearings is from ^^ to J that of plain bearings at moderate 
speeds and loads. As the load on the bearing increased, 



802 



APPLIED MECHANICS FOR ENGINEEliS 



the coefficient of friction decreased. Tightening the bear- 
ing was found to increase the friction considerably. 

Tests of the friction of steel 
rollers 1, 2, 3, and 4 in. in diameter 
are reported in the Trans. Am. 
Soc. C. E., August, 1894. The 
rollers were tested between plates 
1| in. thick and 5 in. wide, ar- 
ranged as shown in Fig. 237. 
Tests were made with the plates 
and rollers of cast iron, wrought 
iron, and steel. The friction P' for 

.0063 




Fig. 236 



unit load P was found to be 



V^ 



0120 
for cast-iron rollers and plates,^ — ^-for wrought iron, and 

Vr 



.0073 



► 7" 



for steel, where r represents the radius of the roller 

Vr 

in inches. The rollers were 
turned and the plates planed, ^ 
but neither was polished. 

163. Ball Bearing^s. — For 
high speeds and light or moder- 
ate loads the friction is much W 
reduced by the use of hardened 
steel balls instead of the steel 
rollers. These bearings are now 

used on all classes of machinery, giving a much greater 
efficiency except for heavy loads. The principal objection 
to the ball bearing seems to be due to the fact that there is 




FRICTION 



303 



'^^^ 



O-^ 



.j^ 



so little area of contact between the balls and bearing 
plates. This gives rise to very high stresses over these areas, 
and consequently a considerable 
deformation of the balls. When 
the ball has been changed from its 
spherical form, it is no longer free 
to roll, and the friction increases 
rapidly. Some authorities con- 
sider a load of from 50 to 150 lb. 
sufficient for balls varying in size 
from ^ to ^ inch in diameter. Fig- 
ure 238 illustrates a type of bear- Fig. 238 
ing used for shafts, and Fig. 239 a type used for thrust blocks. 
The conclusions reached by Goodman from a series of 
tests on bicycle bearings (Proc. Inst. C. E., Vol. 89) are 

as follows : 

(1) The coefficie7it of friction 
of hall hearings is constant for 
varying loads^ hence thefrictional 
resistance varies directly as the 
load. 

(2) The friction is U7iaffected 
by a change of temperature. 

The bearings were oiled be- 
fore starting tlie tests. The 
coefficient of friction for ball 
bearings was found to be rather higher than for plain bear- 
ings with bath lubrication, but lower than for ordinary 
lubrication. Ball bearings will also run easily with a less 
supply of oil. The followiuLi^ tal)le gives the results of 




804 



APPLIED MECHANICS FOR ENGINEERS 



tests of ball bearings. The bearings were oiled before start- 
ing, and the tests were run at a temperature of 68° F. 





19 


167 


360 


Load on 
Bkahinu 


Kkvolutions per Min. 


ItEVOLrTIOXS PER MiN. 


PtEVOUTTIONS FEB MiN. 




Coeff. friction 


Friction, lb. 


Coetf. friction 


Friction, lb. 


Coeif. friction 


Friction, lb. 


10 


.0060 


.06 


.0105 


.10 


.0105 


.10 


20 


.0045 


.09 


.0067 


.13 


.0120 


.24 


30 


.0050 


.15 


.0050 


,15 


.0110 


.33 


40 


.0052 


.21 


.0052 


.21 


.0097 


.39 


50 


.0054 


.27 


.0054 


.27 


.0090 


.45 


60 


.0050 


.30 


.0055 


.33 


.0075 


.45 


70 


.0049 


.34 


.0054 


.38 


.0068 


.47 


80 


.0048 


.38 


.0062 


.49 


.0060 


.48 


90 


.0050 


.45 


.0068 


.61 


.0060 


.54 


100 


.0058 


.58 


.0069 


.69 


.0057 


..57 


110 


.0054 


.59 


.0065 


.71 


.0060 


.66 


120 


.0055 


.66 


.0075 


.90 


.0057 


.68 


130 


.0058 


.75 


.0078 


1.01 


.0062 


.81 


140 


.0056 


.78 


.0077 


1.08 


.0060 


.84 


150 


.0060 


.90 


.0083 


1.24 


.0062 


.93 


160 


.0075 


1.20 


.0081 


1.29 


.0058 


.93 


170 


.0079 


1.34 


.0078 


1.33 


.0055 


.93 


180 


.0079 


1.42 


.0078 


1.40 


.0053 


.95 


190 


.0087 


1.65 


.0076 


1.44 


.0054 


1.03 


200 


.0090 


1.80 


.0081 


1.62 


.0060 


1.20 



Another series of tests, run with a constant load on the 
bearing of 200 lb. and a temperature of 86° F., shows the 
variation of the coei^icient of friction with the speed. It 
is seen that as the speed increased the coefficient and the 
friction decreased. The preceding table, however, shows, 
for loads below 175 lb., an increase in the coefficient with 
increase in speed. In particular, this table shows that for 



FRICTION 



305 



loads below 80 lb. the coefficient increased with increase 
of speed ; for loads between 90 and 175 lb. it increased 
when the speed was 150 r.p.m. and decreased when it was 
350 r.p.m. Beyond 175 lb. the coefficient increased. 



Kevolutions per Minute 


Coefficient Friction 


Friction Pounds 


15 


.00735 


1.47 


93 


.00465 


.93 


175 


.00375 


.75 


204 


.00345 


.69 


280 


.00300 


.60 



It seems from the data given that the first conclusion 
of Goodman's should be changed to read : the coefficient of 
friction of hall hearings is constant for varying loads^ up to 
a certain limits heyond which it increases with increase of 
load. This limit is about 150 lb. in the tests reported. 

Tests on ball bearings designed for machinery subjected 
to heavy pressures have been made in Germany. (See Zeit- 
schrift des Vereins deutsche Ingenieure, 1901, p. 73.) It 
was found that at speeds varying from 65 to 780 revolu- 
tions per minute, where the bearing was under pressures 
varying from 2200 lb. to 6600 lb., the coefficient of friction 
varied little and averaged .0015. 

Tests of ball bearings made by Stribeck and reported by 
Hess (Trans. Am. Soc. M. E., Vol. 28, 1907) give rise to 
the following conclusions : (a) tlie load that may be put 
upon a bearing is given by the formula 

cdhi 



r = 



11.02' 



306 



APPLIED MECHANICS FOR ENGINEERS 



where P is the load in pounds on a bearing, consisting of 
one row of balls, c is a constant dependent upon the mate- 
rial of the balls and supporting surfaces and determined 
experimentally, d the diameter of the balls, the unit being 
I of an inch, and 7^ the number of balls. For modern 
materials c varies from 5 to 7.5. (5) The coefficient of 
friction varied from .0011 to .0095. It was independent 
of speed, "within wide limits," and approximated .0015; 
this was increased to .003 when the load was about one 
tenth the maximum. 

The following values for the coefficient of friction for 
heavy loads are reported, from observation, with the state- 
ment that the real values are probably somewhat less : 



Revolutions 
per minute 


65 


100 


190 


380 


580 


780 


1150 


Coefficient of 
















friction for 
















load 840 lb. 


.0095 


.0095 


.0093 


.0088 


.0085 




.0074 


Coefficient of 
















friction for 
















load 2400 lb. 


.0065 


.0062 


.0058 


.0053 


.0050 


.0049 


.0047 


Coefficient of 
















friction for 
















load 4000 to 
















92.30 lb. 


.0055 


.0054 


.0050 


.0050 


.0041 


.0041 


.0040 



It should be remembered that the friction of a ball 
bearing is due to both sliding and rolling friction, the 
sliding friction being due to the elasticity of the balls 
and the bearing. (See Art. 159.) Rolling friction is most 



FRICTION 307 

nearly approached wlien the balls are hard and not easily 
changed from their spherical shape. All materials, how- 
ever, are deformed under pressure so that perfect rolling 
friction is impossible. On account of the sliding friction 
present in roller and ball bearings, it is necessary to use a 
lubricant to prevent wear. 

Problem 383. IIow many f-in. balls will be necessary in a ball 
bearing designed to carry -iOOO lb., if c = 7.5 ? If /= .0015, what 
work is lost per revolution, the distance from the axis of rotation to 
the center of balls being one inch ? 

164. Friction Gears. — In the friction gears the driver 
is usually the smaller Avheel, and when there is any differ- 
ence in the materials of which the wheels are made, the 
driver is made of the softer material. This latter arrange- 
ment is resorted to, to prevent flat places being worn on 
either wheel in case of slipping. These gears have been 
used for transmitting light roads at high speeds, where 
toothed gears would be very noisy, or in cases where it is 
necessary to change the speed or direction of the motion 
quickly. 

The use of paper drivers has made possible the trans- 
mission of much heavier loads by means of such gears. 
A series of tests, made by W. F. M. Goss, and reported 
in Trans. Am. Soc. M. E., Vol. 18, on the friction be- 
tween paper drivers and cast-iron followers, is of interest 
in this connection. The apparatus used is shoAvn in Fig. 
240. The pressure between the wheels was obtained by 
a mechanism that forced the two wheels together with a 
pressure P. A brake wheel shown in the figure absorbed 
the power transmitted. 



308 



APPLIED MECHANICS FOR ENGINEERS 



The coefficient of friction was regarded as the ratio of 
JP to P, as in sliding friction. While tliis is customary, 
it is not entirely true, since we have the rolling of one 



IRON FOLLOWER 




Fig. 240 

body over the other. We shall, however, assume that we 
may call the coefficient of friction /= —• It was found 

that the coefficient of friction varied with the slippage, 
but was fairly constant for all pressures up to some point 
between 150 to 200 lb. per inch of width of wheel face. 
'^ Variations in the peripheral speed hetiveen 400 and 2800 
ft. per minute do not affect the coefficient of friction.^' 

If the allowable coefficient of friction be taken as .20, 
the horse power transmitted per inch of width of face of 
the wheel, for a pressure of 150 lb., is 

H.P. = ^^^ ^ '^ x-^,7rdxw xJSr ^ .000238 dwN, 
33,000 



FRICTION 



309 



where d is the diameter of the friction wheel in inches, 
w the width of its face in inches, and iV the number of 
revolutions per minute. Using this formula, the following 
table is given in the article in question: 

Horse Power which may he transmitted by Means of Paper 

Friction Wheel of One Inch Face, when run 

rxPER a Pressure of 150 Lb. 



Diameter 






1 


lEvoi.iTiuxs PEU Minute 






OF PlLI.EY 


















IN Inches 


35 


50 


75 


100 


150 


200 


600 


1000 


8 


.047G 


.0952 


.1428 


.1904 


.2856 


.3808 


1.1424 


1.904 


10 


.0595 


.1190 


.1785 


.2380 


.3.570 


.4760 


1.4280 


2.380 


14 


.0833 


.1666 


.2499 


.3332 


.4998 


.6664 


1.9992 


3.332 


16 


.0952 


.1904 


.2856 


.3808 


.5712 


.7616 


2.2848 


3.808 


18 


.1071 


.2142 


.3213 


.4281 


.6426 


.8.568 


2.5704 


4.288 


24 


.1428 


.2856 


.4284 


.5712 


.8568 


1.1424 


3.4272 


5.712 


30 


.1785 


.3570 


.5355 


.7140 


1.0710 


1.4280 


4.2840 


7.140 


36 


.2142 


.4284 


.6426 


.8568 


1.2852 


1.7136 


5.1408 


8.560 


42 


.2499 


.4998 


.7497 


.9996 


1.4994 


1.9992 


5.9976 


9.996 


48 


.2856 


.5712 


.8568 


1.1424 


1.7136 


2.2848 


6.8544 


11.420 



The value of the coefficient of friction for friction 
gears (Kent, " Pocket- 
Book ") may be taken 
from .15 to .20 for 
metal on metal ; .25 to 
.30 for wood on metal ; 
.20 for wood on com- 
pressed paper. 

Problem 384. If the 
friction wheels are grooverl 
as shown in Fig. 241, both Fia. 241 




310 



APPLIED MECHANICS FOR ENGINEERS 



of cast iron, and the small driver fits into the groove of the larger 
follower, prove that the force transmitted is 



F=2fN 



sm a 



Problem 385. The speed of the rim of two grooved friction 
wheels is 400 ft. per minnte. If u = 45°, /= .18, what must be the 
pressure P to transmit 100 horse power? 

Problem 386. What horse power may be tiansmittedby the gear- 
ing in the preceding problem, if P = 0000 lb. and the peripheral 
velocity is 12 ft. per second ? 

165. Friction of Belts. — When a belt or cord passes 
over a pulley and is acted upon by tensions 2\ and T^, the 
tensions are unequal, due to the friction of the pulley on 
the belt. We shall determine the relation between T^ and 
T^' Let the pulley be represented in Fig. 242. The belt 



'\ dfj 




Fig. 242 



covers an arc of the pulley whose angle is a. Consider 
the forces acting upon the belt and suppose T^ and T^ to 
be tlie tensions in tlie belt on the tiofht and slack sides, 



FRICTION 



311 



respectively, and 2^ the tension in the belt at any point of 
the arc of contact. Consider the forces acting on a por- 
tion of the belt of length As = rA/3 (Fig. 242). These 
forces are T, 2^4- A 2^, and Ai^, tangent to the arc, and AP, 
the normal pressure of the pulley on the belt, which may 
be regarded as acting at the center of the arc. Let m be 
the mass of a unit length of the belt. The length of the 
portion considered is then mrA/S. 

Suppose the belt to have uniform speed, v. The forces 
along the tangent will then balance, i.e. 

T+AF= T-hAT. 

.-. dF=dT. 

The acceleration toward the center is — , and the force 

toward the center is 

A/3 



(T+A7+ T) sin^-AP. 

2 



.-. (2 T-^AT)sm 



A^ 
2 



AP = nn'^A/3. 



Dividing by A/3 and passing to the limit, remembering 

A/3] 

= 1, we have, 



that 



lim 



sin 



A^ 
T- 



dP 

d^ 



mv- 



If /is the coefficient of friction, and the belt is on the 
point of slipping, 

- fdp. 



or 



dF 



dP = \dF=\dT. 



f 



f 



ol2 APPLIED MECHAXTCS FOR ENGTNEERS 

Substituting this value of dP^ we have 

fdl3 
dT 
T — mv^ 
Integrating, 



=/'//3. 



logg(7^— mv^^ 



Ti 



=//3 



or loge 7^ =/« 



^2 — ^?'2 

' Ti— niv- 

For low velocities the term mv^ is small compared to T^ 
and 2^2' ^^^ ^1^6 formula may be written 

It should be noted that m in the above formula is the 
mass of a portion of the belt 1 foot long, and that v must 
be reckoned in feet per second if y is taken as 32.2. 

Problem 387. A rope makes two complete turns around a post 
6 in. in diameter. What maximum tension could be balanced by a 
force of 100 lb. if/= .3? 

Problem 388. A weight of 500 lb. is to be lowered by a rope 
wound round a horizontal drum. If the arc of contact is 450° and the 
coefficient of friction is .25, what force is necessary to lower the weight 
uniformly ? 

Problem 389. The velocity of a belt is 3000 ft./min., the tension 
in the tight side is 150 lb. per inch of width, the coefficient of friction 
is .25, and the weight of a portion of the belt 1 ft. long and 1 in. 
wide is .15 lb. If the belt is on the point of slipping and the arc of 
contact is 150°, what is the tension in the slack side? 



FRICTION 313 

Problem 390. A rope is wrapped four times around a post and a 
man exerts a pull of 50 lb. on one end. If the coeHicient of friction is 
.3, what force can be exerted upon a boat attached to the other end 
of the rope ? 

166. Power Transmitted by a Belt. — From the relation 

dF=-dT (Art. 165.) 



there follows 



dT= I dF, 



or T^— T^ = jP, the total friction. 

The work done per second against F is 

Fv or (7\- T^)v. 

Hence the horse power transmitted by the belt is 

HP - (Ti-T,)v 
650 ' 

where T^ and jPg are in pounds and v is in feet per second. 

Problem 391. Show that the formula for H.P. transmitted by the 
belt may be reduced to the form 

' ' 550 

Problem 392. Given a maximum allowable tension, T'j, show that 
the power transmitted is a maximum when 

Problem 393. A pulley 4 ft. in diameter making 200 r. p. m. 
drives a belt that absorbs 20 H.P. The belt is \ in. thick and weighs 
56 lb. per cubic foot. If a = tt and/ = .27, how wide must the belt be 
that the tension may not exceed 75 lb. per inch of width ? 



314 



APPLIED MECHANICS FOR ENGINEERS 



Problem 394. What II. P. may be transmitted by a belt 6 in. 
wide, \ in. thick, weighing 56 lb. per cubic foot, when traveling at 
1500 ft. per minute, if (t = 108°,/= .25, and the maximum tension is 
300 lb. per square inch ? 

Problem 395. Find the maximum H. P. that can be transmitted 
by the belt in the preceding j^roblem, and the corresponding velocity. 



167. Transmission Dynamometer. — It has been shown, 

in Art. 165, that the tension of a 
belt on the tight side is greater 
than the tension on the slack side. 
The transmission dynamometer 
(the Fronde dynamometer), illus- 
trated in Fig. 243, is designed to 
measure tlie difference in these 
tensions. Let the pulley D be the 
T^ driver and the pulley U the fol- 
lower, so that T^ represents the 
tight side of the belt and T^ the 
slack side. The pulleys B^ B run 
loose on the T-shaped frame CBB. 
This frame is pivoted at A. If we 
neglect the friction due to the loose 
pulleys, we have the following 
forces acting on the T-frame, two 
forces T^ at the center of the right- 
liand pulley B^ two forces T^ at 
the center of the left-hand pulley 
B^ a measurable reaction P at (7, 
Taking moments about 





Fig. 243 



and tlie reaction of the pin at A. 
the pin, we have 



FRICTION 315 

P( CA) = 2 T^iBA) - 2 T^iBA) 
= 2BA(T,-T,-), 

so that T^- To = ^%^' 

The distances CA and ^^ are known, and P may be 
measured ; the difference, then, T^^ — T^^ may always be 
obtained. The value T-^ — T^ is then known and the horse 
power determined by the relation 

^'^'~ 33;000 ' 

where 7i is the number of revolutions per minute, and r is 
the radius of the machine pulley in feet. 

168. Creeping or Slip of Belts. — A belt that transmits 
power between two pulleys is tighter on the driving side 
than it is on the following side. On account of this differ- 
ence in tension and the elasticity of the material, the tight 
side is stretched more than the slack side. To compen- 
sate for this greater stretch on one side than on the other, 
the belt creeps or slips over the pulleys. This slip has 
been found for ordinary conditions to vary from 3 to 12 
ft. per minute. The coefficient of friction when tlie slip 
is considered is about .27 (Lanza). It has also been 
found that the loss in horse power in well-designed belt 
drives, due to slip, does not exceed 3 or 4 per cent of the 
gross power transmitted, and that ropes are practically 
as efficient as belts in this respect. For an account of the 
experimental investigations on this subject the student is 
referred to Inst. Mecli. Eng., 1895, Vols. 3-4, p. 599, and 
Trans. Am. Soc. M. E., Vol. 26, 1905, p. 584. 



316 



APPLIED MECHANICS FOR ENGINEERS 



169. Coefficient of Friction of Belting. — The value of the 
coefficient of friction of belting depends not only on the 
slip bat also upon the condition and material of the rubbing 
surfaces. Morin found for leather belts on iron pulleys the 
coefficient of friction /= .56 wlien dry, .36 when wet, .23 
when greasy, and .15 when oily (Kent, "Pocket-Book"). 
Most investigators, however, including Morin, took no ac- 
count of slip, so that the best value of /, everything con- 
sidered, is that given in the preceding article (.27). 

170. Friction of a Worn Bearing. — The friction of a 
bearing that fits perfectly is the friction of one surface 
sliding over another and is given by the equation 

F=fN, 

where F is the force of friction, / the coefficient of friction, 
and iVis the total normal pressure on the bearing. 

When, however, the bearing is worn, as is shown much 
exaggerated in Fig. 244, the friction may be somewhat 

B 




Fig. 2M 



FRICTION 317 

different. When motion begins, the shaft will roll up on 
the bearing until it reaches a point A where slipping be- 
gins. If motion continues, slipping will continue along a 
line of contact through A. Let P be a force that causes 
the rotation, R a force tending to resist the rotation, and 
J?j the reaction of the bearing on the shaft. There are 
only three forces acting on the shaft, so that P, i2, and R^ 
must meet in the point B. The direction of R^ is accord- 
ingly determined. The normal pressure is N= R^ cos ^, 
and the force of friction is 

It is seen that 6 is the angle of friction. The moment of 
the friction with respect to the center of the axle is 

Fr = JRxV sin 9. 

If the axle is well lubricated, so that 6 is small and sin 6 
may be replaced by tan 6 =f^ the friction is 

F = flt„ 

and the moment 

Fr = fRxr, 

The circle tangent to AB of radius r sin 6 is called the 
friction circle. Since r and 6 are known generally, this 
circle may be made use of in locating the point A, 

The shaft will continue to rotate in the bearing so long 
as the reaction R^ falls within the friction circle, and 
slipping will begin as soon as the direction of Pj becomes 
tangent to the friction circle. 

Problem 396. If the radius of the shaft is 1 in.. 6 = 4°, 
P = 500 lb., Oj = 3 ft., ag = 2 ft., angle between ai and 02 is 100'^, and 



318 



APPLIED MECHANICS FOR ENGINEERS 



P aud R are right angles to n^ and a.^, what resistance R may be 
overcome by P when slipping occurs ? 

Problem 397. The radius of a shaft is 1 in., A' = 20 lb., 
P = 20 lb., r/j = 3 ft., and «., = 2 ft. What force of friction will be 
acting at the point A, when the angles between P and ai and R and 

a 2 are right angles? What must be the 
value of the coefficient of friction? 



171. Friction of Pivots. — The 

friction of pivots presents a case of 

sliding friction, so that tlie force of 

friction F equals the coefficient of 

friction times the normal pressure. 

That is, 

r = fN. 

(a) Flat- end Pivot. Assuming 
the pivot to press uniformly on 
the bearing, the friction on an 
element of area r'dOdr' is 



-<- 



->- 




f—r'dedr'. (Fig. 245.) 
The total work done agrainst fric- 



FiG. 245 tion per revolution is 

rr /»2,r p 

Work per rev. = 1)2 irr'f — - r'dr'du 

%Jo «/ ITT 

= ^ r Cr"'dr'dd, 

7-2 Jo Jo . 



or 



4 irrfP 
Work per revolution = — e* — 

o 



(ft) Collar Bearing or Hollow Pivot. 



FRICTION 



319 



Here the work per revolution becomes (Fig. 246) 
Work per rerolntioii = f *^' T"^ 2 -nr' --4- — ^ r'dr'dQ 



_ 4_Tr ?•:] — ^'i 
~ X r1 - ?i 



/i* 




1^ ^ 

Fig. 246 



(c) Conical Pivot. The conical pivots, illustrated in 
Fig. 247, do not usually fit into the step the entire depth 
of the cone. Let the radius of the cone at the top of the 
step be rj, a half the angle of the cone, and let dP^ be the 
normal pressure of the bearing on an elementary area whose 
horizontal projection is r'dr'dO. 



820 



APTLIED MECHANICS FOR ENGINEERS 



The vertical com- 
ponent of c?Pj is 
equal to the vertical 
load on the liori- 
zontal area r'dr'dO 
which is 

dP = -^r'dr'de. 

TTVi 

Trrf sin a 

and the force of fric- 
tion acting on this 
element is 

F.G. 247 '^'■1 «'" « 

The total work per 

revolution done against friction is therefore 

Work per revolution = C^ C — f^^ — 2 irr'^dr'dQ 

Jo Jo irr'i Sin a • 

_ 4 Trn/P 

3 sin a 




TT 



If a = ^i this value for work lost reduces to the work 

lost per revolution, in the case of the flat-end solid pivot. 
It is easily seen, since sin a is less than unity, that if r^ is 
nearly equal to r, the friction of the conical bearing is 
greater than tlie friction of the flat-end bearing. Tliis 
might have been expected from the wedgelike action of 
the pivot on the step. It is also easily seen that r^ may 



FRICTION 



321 



be taken small enough so that the friction will be less 
than the friction of the flat pivot. The work lost due to 
friction in the case of the conical pivot will be equal to, 
greater, or less than, the work lost, due to friction in the 
case of the flat-end pivot, 
according as 



r, > r sin a. 

< 

(c?) Spherical Pivot. 
Suppose the end of the 
pivot is a spherical sur- 
face, as shown in Fig. 
248. Let r be the radius 
of the shaft and r^ the 
radius of the spherical 
surface ; then the load 
per unit of area of 

P_ 



horizontal surface is 




Fig. 248 



Trr" 



The horizontal projection of any elementary ring of the 
bearing, of radius x^ is 2 irxdx. Tlie load on this area is 

and the corresponding normal pressure is 

jn 2 Pxdx r, 

dP. = — sec p, 

and the total work done asfainst friction in one revolution 



IS 



Jo 



TrfPx^dx 
'o r^ cos yS 



322 APPLIED MECHANICS FOR ENGINEERS 

Here /S is a variable such that 

2: = rj sin yS. 
.*. dx = r^ cos fid/3, 

and the expression for the work becomes 

Work per rev. = — ^- — 1 I sui'' pap 

4 7rfP)ifa 1 . 
= — *^ ^1 77 — ^ sm a cos a 



= 2 ir/Pri r^-5: COtal, 

Lsm-a J 



since r = r^ sm a. 



IT 



If the bearing is hemispherical, a = — , and the work 
lost per revolution becomes 

The friction of flat pivots is often made much less by 
forcing oil into the bearing, so that the shaft runs on a 
film of oil. In the case of the turbine shafts of the Niag- 
ara Falls Power Company (see Art. 143) the downward 
pressure is counteracted by an upward water pressure. 
In some cases the end of a flat pivot has been floated on 
a mercury bath. This reduces the friction to a minimum. 
(See Engineering, July 4, 1893.) 

The Schiele pivot is a pivot designed to wear uniformly 
all over its surface. The surface is a tractrix of revolu- 
tion ; that is, the surface formed by revolving a tractrix 
about its asymptote. Its value as a thrust bearing is 
not as great as was first anticipated. (See American Ma- 
chinist, April 19, 1894.) 



FRICTION 323 

The coefficient of friction for well-lubricated bearings 
of flat-end pivots has been found to vary from .0044 to 
.0221. (See Proc. Inst. M. E., 1891.) For poorly lubri- 
cated bearings the coefficient may be as high as .10 or 
.25 for dry bearings. 

Problem 398. Show that the work lost per revolution for the 
hemispherical pivot is 2.35 times the work lost per revolution for the 
flat pivot. 

Problem 399. The entire weight of the shaft and rotating parts 
of the turbines of the Niagara Falls power plant is 152,000 lb., the 
diameter of the shaft 11 in. If the coefficient of friction is con- 
sidered as .02 and the bearing a flat-end pivot, what work would be 
lost per revolution due to friction ? 

Problem 400. A vertical shaft carrying 20 tons revolves at a 
speed of 50 revolutions per minute. The shaft is 8 in. in diameter 
and the coefficient of friction, considering medium lubrication, is 
.08. What work is lost per revolution if the pivot is flat? What 
horse power is lost ? What horse power if the pivot is hemispherical? 

Problemi 401. What horse power would be lost if the shaft in 
the preceding problem was provided with a collar bearing 18 in. out- 
side diameter instead of a flat-end block? Compare results. 

Problem 402. A vertical shaft making 200 revolutions per minute 
carries a load of 20 tons. The shaft is 6 in. in diameter and is 
provided with a flat-end bearing, well lubricated. If the coefficient 
of friction is .004, what horse power is lost due to friction? 

172. Absorption Dynamometer. — The friction brake shown 
in Fig. 240 is used to absorb the energy of the mechanism. 
It may be used as a means of measuring the energy, and 
when so used it may be called an absorption dynamometer. 
The weight TF, attached to one end of the friction band, 
corresponds to the tension in the tight side of an ordinary 



324 APPLIED MECHANICS FOR ENGINEERS 

belt (see Art. 165), while the force measured by the spring 
S corresponds to the tension on slack side of a belt. Let 
W= T^ and S= T^\ then T^ — T^e^°-, just as was found in 
the case of belt tension. The work absorbed per revolu- 
tion is work = (I'j — 2^2)2 Trr-^, where r^ is the radius of the 
brake wheel. The horse power absorbed is 

33,000 

where n is the number of revolutions per minute. 

In many cases the friction band is a hemp rope, and in 
such cases it is possible to wrap the rope one or more times 
around the pulley, making it possible to make T^ — T^ 
large while T^ is small. 

The surface of the brake wheel may be kept cool by allow- 
ing water to flow over the inside surface of the rim, which 
should be provided with inside flanges for that purpose. 

173. Friction Brake. — The friction brake shown in 
Fig. 249 consists of the lever EC^ the friction band, and 
the friction wlieel. Such brakes are used on many types 
of hoisting drums, automobiles, etc. Let the band ten- 
sions be T^ and T^^ and let W be the force causing the 
motion, that is, the working force, and P the force applied 
at the end of the lever EQ in such a way as to retard the 
rotation of the drum. We have here as before T^ = Tc^e^°- 
and the work per revolution {T^— T,^^iTr^-\- F^irr^,- 
By taking moments about A we have for uniform motion 

T^ — T^= 1^^^ 3^ where r^ is the radius of the shaft 

and F is the force of friction acting on the shaft. Taking 



FRICTION 



325 



moments about (7, we have 

P{EC) = T^d^ sin S + T^d^ sin /9. 




Fig. 249 

Problem 403. A weight of one ton is being lowered into a mine 
by means of a friction brake. The radius of the drum is 1^ ft., 
radius of the friction wheel 2 ft., coefficient of brake friction .30, 
8 = 4.5^ /? = 15S r/, = r/^ = 1 ft., EC = ^ ft., radius of shaft 1 in., 
coefficient of axle friction -Oi, and the weight of the drum and brake 
wheel is 600 lb. Find T^, T^, and P in order that the weight W 
may be lowered with uniform velocity. 

Problem 404. Suppose the weight in the above problem is being 
lowered with a velocity of 10 ft. per second when it is discovered that 
the velocity must be reduced one half while it is being lowered the 
next 10 ft., what pressure P will it be necessary to apply to the lever 
at £J to make the change? What will be the tension in the friction 
bands and the tension in the rope that supports PV? 

Problem 405. If the weight in the above problem has a velocity 
of 10 ft. per second, and it is required that the mechanism be so con- 
structed that it could be stopped in a distance of 6 ft . what pressure 
P on the lever and tensions I\ and T"., would it require? What 



326 



APPLIED MECHANICS FOR ENGINEERS 



would be the tension in the rope caused by the sudden stop? Cora- 
pare this tension with W, the tension when the motion is uniform. 

174. Prony Friction Brake. — The Pron}- friction brake 
may be used as an absorption dynamometer as shown in 
principle in Fig. 250. Let W be a working force acting 
on the wheel of radius r and suppose the brake wheel to 




Fig. 250 

be of radius r^. The brake consists of a series of blocks 
of wood attached to the inner side of a metal band in 
such a way that it may be tightened around the brake 
wheel as desired by a screw at B. This band is kept from 
turning by a lever OAD, held in the position shown by an 
upward pressure P, at A. Considering the forces acting 
on the brake and taking moments about the center, we 
have the couple due to friction, Prj, equal to the moment 
PiOA),ov ,Fr,^p(^OA~). 

Considering the forces acting on the wheel, and neglecting 
axle friction, we get for uniform motion of TF, 

Fr^^Wr. 



FRICTION 



827 



The energy absorbed is used in heating the brake wheel. 
The wheel is kept cool by water on the inside of the rim. 
The work absorbed per minute is 2 irr^Fn = 2 7rP( OA)n^ 
where n is the number of revolutions per minute. The 
force P may be measured by allowing a projection of the 
arm at A to press upon a platform scales. The horse 
power absorbed is 

^'^'- 33,000 ' 

where OA is expressed in feet, and n is the number of 

revolutions per minute. 

33 
If OA be taken as — -^, a convenient length, the formula 

2 TT 

reduces to 



H.P. = 



1000 



A dynamometer slightly different from the Prony dy- 
namometer is shown in Fig. 251. It differs only in the 

In this case the force P is meas- 



means of measuring P. 




Fui. 251 



328 APPLIED MECHANICS FOR ENGINEERS 

ured by the angular displacement of a heavy pendulum 
W\. Taking moments about the axis of W^ and calling 
/•g the distance from that axis to its center of gravity and 
fi the angular displacement, we have 

Fr^ = TTi^g sin jS, 

so that the horse power absorbed may be written 

^•^•" ^4 33,000 ■' 

where OA^ r^, and r^ are expressed in feet. If OA be 

taken as , this becomes 

27r' 

Wiv^ sin pn 



H.P. = 



n 1000 



The student should understand that the rotation of the 
mechanism at is not in every case due to a weight W 
being acted upon by gravity. In fact, in most cases, the 
motion will be due to the action of some kind of engine. 
This, however, will not change the expressions for horse 
power. 

QQ 

Problem 406. If W^ = 100 lb., OA =^--,r^ = 2 ft., and r^ = Q 

in., what horse power is absorbed by the brake if y8 is 30"^, and n is 
300 revolutions per minute ? 

175. Friction of Brake Shoes. — The application of the 
brake shoe to the wheel of an ordinary railway car is 
shown in Fig. 252, where W is the axle friction, F the 
brake-shoe friction, iV the normal pressure of the brake 
shoe, Gr the weight on the axle, and F^ and N^ the reaction 



FRICTION 



329 




Fig. 252 



of the rail on the wheeL The brakes on a railway car 
when applied should be capable of absorbing all the en- 
ergy of the car in a 
very short time. The 
high speeds of modern 
trains require a system 
of perfectly working 
brakes, capable of stop- 
ping the car when 
running at its maxi- 
mum speed in a very 
short distance. 

The coefficient of 
friction between the shoes and wheel for cast-iron wheels 
at a speed of 40 mi. per hour is about ^, while at a point 
15 ft. from stopping the coefficient of friction is increased 
7 per cent, or it is about .27. The coefficient for steel- 
tired wheels at a speed of 6b mi. per hour is .15, and at a 
point 15 ft. from stopping it is .10. (See Proc. M. C. B. 
Assoc, Vol. 39, 1905, p. 431.) 

The brake shoes act most efficiently when the force of 
friction F is as large as it can be made without causing a 
slipping of the wheel on the rail (skidding). The normal 
pressure iV, corresponding to the values of the coefficient 
of friction given above, varies in brake-shoe tests from 
2800 lb. to 6800 lb., sometimes being as high as 10,000 lb. 

Problem 407. A 20-ton car moving on a level track with a 
velocity of a mile a minute is subjected to a normal brake-shoe pres- 
sure of 6000 lb. on each of the 8 wheels. If the coefficient of brake 
friction is .15^ how far will the car move before coming to rest? 



330 APPLIED MECHANICS FOR ENGINEERS 

Problem 408. In the above problem the kinetic energy of rota- 
tion of the wheels, the axle friction, and the rolling friction have been 
neglected. The coefficient of friction for the journals is .002, that 
for rolling friction is .02. Each pair of v^heels and axle has a mass 
of 45 and a moment of inertia with respect to the axis of rotation of 
37. The diameter of the wheels is 32 in. and the radius of the axles 
is 2| in. Compute the distance the car in the preceding problem will 
go before coming to rest. Compare the results. 

Problem 409. A 30-ton car is running at the rate of 70 mi. per 
hour on a level track when the power is turned off and brakes ap- 
plied so that the wheels are just about to slip on the rails. If the 
coefficient of friction of rest between wheels and rails is .20, how 
far will the car go before coming to rest ? 

Problem 410. A 75-ton locomotive going at the rate of 50 mi. 
per hour is to be stopped by brake friction within 2000 ft. If the 
coefficient of friction is .25, what must be the normal brake-shoe 
pressure ? 

Problem 411. A 75-ton locomotive has its entire weight carried 
by five pairs of drivers (radius 3 ft.). The mass of one pair of drivers 
is 271 and the moment of inertia is 1830. If, when moving with a 
velocity of 50 mi. per hour, brakes are applied so that slipping on 
the rails is impending, how far will it go before being stopped ? The 
coefficient of friction between the wheels and rails is .20. 

176. Train Resistance. — The resistance offered by a train 
depends upon a number of conditions, sucli as velocity, 
acceleration, the condition of track, number of cars, curves, 
resistance of the air, and grades. No law of resistance 
can be worked out from a theoretical consideration, be- 
cause of the uncertainty of the influence of the various 
factors involved. Formulse have been developed from the 
results of tests ; the most important of these are given 
below. 



FRICTION 331 

Let H represent the resistance in pounds and v the 
velocity in miles per hour. W. F. M. Goss has found that 
the resistance may be expressed as 

B= .0003(Z + 347)?;2, 

where L is the length of the train in feet. (See Engineer- 
ing Record^ May 25, 1907.) 

The Baldwin Locomotive Works have derived the 

formula i? = 3 + - 

6 

as the relation between the resistance and velocity. When 
all factors are considered, this becomes 

i2 = 3 + ^ + .3788(0 + .5682(c?) + .1265(a), 
6 

where t = grade in feet per mile, c the degree of curvature 
of the track, and a the rate of increase of speed in miles 
per hour in a run of one mile. 

To get the total resistance it is necessary to include, in 
addition to the above factors, the friction of the locomo- 
tive and tender. This is given by Holmes (see Kent's 
"Pocket-Book") as 

R^= [12 +.3(v- 10)17], 

where TTis the weight of the engine and tender in pounds 
and R^ the resistance in pounds due to friction. 

Other formulse derived as the result of experiments are 
shown graphically in Fig. 253. 



332 



APPLIED MECHANICS FOR ENGINEERS 




RESISTANCE VELOCITY CURVE^ 
FOR RAILROAD TRAINS I 




10 20 30 40 50 60 70 bO 90 100 

VELOCITY IN IVIILE8 PER HOUR 



Fig. 253 



FRICTION 



333 



The formulse themselves are as follows (see Engineering, 
July 26, 1907) : 



CuKVB Number 


Formula 


Authority 


1 


^ = ^^^1:4 


Clark 


2 


^-^-m 


Clark 


3 


^=^-^^^;62 


Wellington 


•4 


290 


Deeley 


5 


i? = .2497 V 


Laboriette 


6 


R = 3.36 + .1867 v 


Baldwin Company 


7 


R = 4.48 + 284 v 


Lmidie 


8 


i^ = 2 + .24 y 


Sinclair 


9 


i? = 2.5+ ^^ 
65.82 


Aspinal 



It is evident that these formulae do not agree as closely 
as one would wish. The difference must be due chiefly to 
the different conditions under which the tests were made. 
These conditions should be taken into account in any 
application of the formulae to special cases. 



CHAPTER XIV 

DYNAMICS OF RIGID BODIES 

177. Statement of D'Alembert's Principle. — A body may 
be considered as made up of a collection of individual 
particles held together by forces acting between them. 
The motion of a body concerns the motion of its individ- 
ual particles. We have seen that in dealing with such 
problems as the motion of a pendulum it was necessary to 
consider the body as concentrated at its center of gravity ; 
that is, to consider it as a material point. The principle 
due to D'Alembert makes the consideration of the motion 
of bodies an easy matter. Consider a body in motion due 
to the application of certain external forces or impressed 
forces. Instead of thinking of the motion as being pro- 
duced by such impressed forces, imagine the body divided 
into its individual particles and imagine each of the parti- 
cles acted upon by such a force as would give it the same 
motion it has due to the impressed forces. These forces 
acting upon the individual particles are called the effective 
forces. D'Alembert's Principle, then, states that the im- 
pressed forces will he in equilibrium with the reversed effec- 
tive forces. 

This amounts to saying that the system of effective 
forces is equivalent to the system of impressed forces. 
The truth of the principle is evident if it be granted that 

334 



DYNAMICS OF RIGID BODIES 



335 



the forces acting between any two particles of the body 
are equal and opposite and act in the line joining the 
particles. In any summation of all the forces acting on 
all of the particles of the body these forces acting between 
the particles of the body would cancel out and leave the 
summation of the impressed forces. 

178. Simple Translation of a Rigid Body. — A body has 
pure translation when all points of the bod}^ have at each 
instant the same velocity. 

Consider a body (Fig. 
254) under the action of 
the impressed forces Pj, Pg' 
Pg,-', and let a be the ac- 
celeration of all particles of 
the body. Choose axes of 
reference, taking the rc-axis 
parallel to the direction of 
the acceleration. Think of 
the body as divided up into 
particles and each particle 
under the action of the effective forces. Let dM be the 
mass of any particle of the body. The effective forces 
acting on dM can be combined into a single force, <iP, 
parallel to the direction of the acceleration, according to 
Newton's second law, such that 

dF=a 'dM, 

Since the effective forces are parallel and proportional 
to the weights of the elements of mass on which they act, 
their resultant is a force, P, parallel to the direction of the 



^ 






/^ dM 


t 




&^ 


\ 


^^x^^^^^ 


-^ 


a 





Fig. 254 



336 



APPLIED MECHANICS FOR ENGINEERS 



acceleration, through the center of gravity of the hody^ such 
that 

F=^adM, 

or F=Ma. 

Hence the resultant of the impressed forces must be a 
force, M • a, parallel to the direction of the acceleration and 
acting through the center of gravity. 

Therefore, if a body has a motion of pure translation, 
the resultant of all impressed forces acting on the body is 
parallel to the direction of the acceleration and acts 

through the center 
of gravity of the 
body. Also, the 
acceleration of the 
center of gravity 
of the body is the 
same as if the 
whole mass were 
concentrated at the 
center of gravity 
and the resultant 
force acted at that 
point. 

179. Rotation of 
a Body about a 
Fixed Axis. — Let 

Fig. 255 represent a body rotating about a fixed axis, here 
chosen as the axis of 2, under the action of forces Pj, Pg* 




Fig. 255 



DYNAMICS OF RIGID BODIES 337 

Pg, •••. Let a be the angular acceleration of the body at 
any mstant. Think of the body as divided up into infini- 
tesimal elements and let dM be the mass of any such 
element, distant r from the axis of rotation. The effec- 
tive forces acting on this element can be combined into 
two components, dT and c?iV, respectively, along the 
tangent and normal to the circular path, since the motion 
of the particle is in a circle whose plane is perpendicular 
to the axis of rotation. The values of dT -dnd 6?iVare 

dT= atdM = radM, 

dN= r(D^dM. (Arts. 117, 126.) 

Applying D'Alembert's principle, and taking moments 
of reversed effective forces and impressed forces about 
OZ, we get 

2 moms impressed forces — J rt?!^ = 0, 

or 2 mom — | r^adM= 0, 

or 2 mom = a J, 

in which Zis the moment of inertia of the body with re- 
spect to the axis of rotation. 

Therefore, when a body rotates about a fixed axis, the 
sum of the moments of the impressed forces about the 
axis of rotation is equal to the angular acceleration of 
the body times the moment of inertia of the body about 
the axis of rotation. 

180. Equations of Motion of the Rotating Body. — In addi- 
tion to the above equation there are five other equations 
which express the conditions of equilibrium of the im- 



338 



APPLIED MECHANICS FOR ENGINEERS 




Fig. 256 



pressed forces and the reversed effective forces (Art. 177). 

The six conditions may be stated as follows : The sum 

of the components of the 
impressed forces parallel to 
each of the three axes equals 
the sum of the components 
of the effective forces par- 
allel to those axes, and the 
sum of the moments of the 
impressed forces about each 
of the axes equals the sum 
of the moments of the effec- 
tive forces about those axes. 
Representing impressed forces with a subscript, z, and 

reckoning moments in the direction shown in Fig. 256, 

these six conditions may be written, 

^Xi = - fcos (/) ^iV- fsin cj>dT, 
2 Fi = - jsin ^dN-\- fcos c^xJiT, 

2 mom^vp = I 2 sin (j)dN— | z cos <^dT^ 
2 vciomiy — — i z cos (f>dN— J z sin (\idT^ 
]Smom,-2 = I rdT. 

Replacing dN and dT by their values from Art. 179, 
and replacing r cos (j> and r sin respectively by x and y^ 
these equations reduce to the forms, 



DYNAMICS OF RIGID BODIES 339 

2Xi = - co^xM - ayM, (1) 

2 r;. = - ay^yM + axM, (2) 

tZ,= Q; (3) 

2 mom,,, = ©2 I yzdM — « j xzdM^ (4) 

2 mom^y = — 0)2 I xzdM— a J yzdM^ (5) 

2 mom^-2 = ai^. (6) 

These equations hold true at any instant of the motion of 
the body. Since x and y are the coordinates of the center 
of gravity, if the axis of rotation passes through the 
center of gravity, the right-hand sides of equations (1) 
and (2) reduce to zero. 

Also if the a;^-plane is a plane of symmetry of the body, 

I xzdM and I yzdM both reduce to zero, since for every 

term xQ-\- z)dM there is a corresponding term xQ— z~)dM, 
and for every term y(^-\-z~)dM there is a corresponding 
term ?/(— z')dM. 

Therefore, when the axis of rotation passes through 
the center of gravity, and is taken as the 2-axis, and the 
rr^-plane is a plane of symmetry of the body, the six 
equations for the forces acting on the body become : 

2^.= 0, (10 

SF, = 0, (20 

2Z,= 0; (3') 

5:mom,,= 0, (4') 

S mom.y = 0, (50 

2 mom^, = al^. (6') 



340 



APPLIED MECHANICS FOR ENGINEERS 



This case is the one that usually comes up in engineer- 
ing problems, and so these simplified equations are more 
often used than the six more general equations. It will 
be noticed that these equations are exactly the same as 
the conditions for equilibrium as determined in Art. 56, 
except that 2 mom^-^ is not zero. 

181. Reaction of Supports of a Rotating Body. — When 
a body is rotating about a fixed axis, the reactions of the 
supports on the body are impressed forces and the appli- 
cations of equations (1) to (6) of the preceding article will, 

when the other impressed 
forces and the angular veloc- 
ity are given, enable one to 
find the reactions of the sup- 
ports. As an illustration 
suppose the body (Fig. 257) 
to be a cast-iron sphere, radius 
2 in., connected to a vertical 
axis by a weightless arm 
whose length is 4 in., the 
^ axis being supported at two 
points as shown. The reac- 
tions of the lower support 
are P^, Fy, P^, and of the 
upper Pi, P;, 0. Let the 
body be rotated by a cord 
running over a pulley of 
radius 1 in. situated 2 in. 
below P^. Let the constant tension, P, in the cord be 10 
lb. and suppose it acts parallel to the ?/-axis. Consider 




Fig. 257 



DYNAMICS OF RIGID BODIES 341 

the motion when the sphere is in the xz-^Vdiie. Take the 
icy-plane through the center of the sphere perpendicular 

to 2J, then j xzdM and j yzdM are both zero. Using the 

foot-pound-second system of units, and the weight of cast 
iron as 450 lb. per cubic foot, we have ^ = |, ^ = 0, 
a = 8.727 lb., M= .271, and 7, = .0708. 
Equations (1) to (6) (Art. 180) then become, 

P, + Pi = -. 1365 0.2, (1) 





P^ + P;-10= .1855 a, 


(2) 




P,- 8.727 = 0, 


(3) 




^•^-p;+ii', = o. 


(4) 


P' - 


-ii'. + K8-727) = 0, 


(5) 




if=.0708«. 


(6) 



From (6) we obtain 

a= 11.77 rad./sec.2. 

Suppose the body to begin to rotate from rest and at 
the time under consideration it has been rotating 2 seconds. 

Then « = 2 a = 23.54 rad./sec. 

Solving the remaining equations, we obtain, 

P, = - 47.15 lb. Pi= - 27.94 lb. 

P, = 2.17 1b. PJ=9.42 1b. 

P, = 8.73 1b. 

A negative sign indicates that the force acts opposite to 
the direction assumed in the figure. 



342 



APPLIED MECHANICS FOR ENGINEERS 




Problem 412. A flywheel 3 ft. in diameter. The cross section of 
the rim is 3 in. x 3 in. and is made of cast iron. Neglect the weight 
of the spokes. This wheel is placed on the axis of Fig. 257, with its 
center on the axis, instead of the sphere. If the other conditions are 
the same, find the reactions of the supports. 

Problem 413. The sphere in Fig. 257 is replaced by a right 
circular cast-iron cone of height one foot and diameter of base one 

foot. The vertex is placed at the point 
of attachment of the sphere, and the base 
is outward. If the other conditions are 
the same, find the reactions of the sup- 
ports. 

Problem 414. In the problem of the 
sphere (Art. 181) find the angular veloc- 
ity at the end of 30 sec. and the reactions 
oi the supports for such speed. 

Problem 415. The two spheres of 
Fig. 258 each weigh 220 lb. and are 12 
in. in diameter. The rod connecting 
them is inclined 30° to the horizontal and 
is rigidly connected to the vertical axle 
AB. Find the reactions at A and B when the system is rotating 
uniformly at 120 r. p. m., neglecting the weight of the rod. 

Problem 416. The 

spheres C and D of Fig. 

259 weigh respectively 100 

lb. and 50 lb. and their 

centers are 2 ft. and 1^ ft. 

respectively from the axis I 

of the shaft. Their planes 

are 90° from each other. 

Find the reactions at the 

supports A and B when 

they are revolving at the rate of 150 r. p. m., the reactions being com- 



FiG. 258 




DYNAMICS OF RIGID BODIES 



343 



puted when C is (a) vertically above the shaft, (&) vertically below 
the shaft. 

Problem 417. A uniform sphere of radius 1 ft. and weight 2000 
lb. is mounted on a horizontal axle of diameter 4 in., round which 
a small cord is wound carrying a weight of 50 lb. Neglecting fric- 
tion, find the angular acceleration of the sphere, the linear accelera- 
tion of the weight, the tension in the cord, and the velocity of the 
weight at the end of 5 sec. 

Suggestion. Consider the motion of the sphere and the weight 
separately, under the action of the forces acting on each. 

Problem 418. A rectangular slab 6 ft. by 8 in. by 2 in. is sup- 
ported by an axis through the slab perpendicular to the face 6 ft. by 
8 in. at a distance of 1 ft. from one end and 4 in. from the side. 
The slab is pulled aside to a horizontal position and released. Find 
the reaction on the sup- 
port when the slab has ^^ ^' 
turned through 60° ; when 
through 90°, the weight of 
the slab being 320 lb. 

SuGGESTioisi". Choose z- 
axis as axis of support and 
a:-axis along the axis of the 
slab at the instants in 
question. 

Problem 419. Two 

drums whose radii are r^ = 

16 in. and r^ = 12 in. are 

mounted as shown in Fig. 

260. Their combined 

weight is 200 lb. and k = 

14". The forces Gi and G2 act upon the drum, as well as journal 

friction amounting to 16 lb. The radius of the shaft is 1 in. Find 

the velocities of G-^ and G2 and the drums when the point of attach- 




FiG. 260 



344 



APPLIED MECHANICS FOR ENGINEERS 



ment of the cord on the small drum has traveled from rest at yl to ^' 
through 90°. Neglect the friction at B. 

182. The Compound Pendulum. — A body rotating under 

the action of gravity about a 
horizontal axis is called a com- 
pound pendulum. 

Let Fig. 261 represent a com- 
pound pendulum rotating about 
an axis through perpendicular 
to the plane of the paper. Let 
the distance from the axis of ro- 
tation to the center of gravity of 
the body be d. 

Applying D'Alembert's princi- 
ple, we have 




Fig. 261 



— Grd sin 6 = aL 



0' 



or, since 



Y ^J' 7,9 

-^0 — -— «^o' 

a= —^ sin 6. 

It was shown in Art. 120 that the tangential acceleration 
of a simple pendulum of length I is 

at = -g sin 6, 
and hence its angular acceleration is y, or 

a = — ^ sin 6. 

C 

Hence the angular motion of a compound pendulum is 
exactly the same as that of a simple pendulum whose 



DYNAMICS OF RIGID BODIES 345 

lengtli, Z, is such that 

or '"d 

7,2 

This value, —?, is called the length of the equivalent simple 
pendulum. 

The time of a small vibration of a compound pendulum 
is therefore 

The point, 0, of the pendulum, about which rotation 
takes place, is called the center of suspension. 

The point O on OC such that 00' =^ is called the 

center of oscillation. It is that point at which the whole 
mass might be concentrated without changing the time of 
vibration, the center of suspension remaining the same. 

Problem 420. A board 4 ft. by 1 ft. by 1 in. vibrates about an 
axis perpendicular to the 4 ft. by 1 ft. face through a point of the 
board 18 in. from the center. Find the time of vibration for small 
oscillations, and the length of the equivalent simple pendulum. 

Problem 421. Show that the time of vibration is the same (for 
equal angles of vibration) for all parallel axes of suspension that are 
at equal distances from the center of gravity of the body. 

Problem 422. A cast-iron sphere w^hose radius is 6 in. vibrates 
as a pendulum about a tangent line as an axis. Find the period 
of vibration and the length of a simple pendulum having the same 
period. Locate the center of oscillation. 



346 APPLIED MECHANICS FOR ENGINEERS 

183. Centers of Oscillation and Suspension Interchangeable. — 

If k is the radius of gyration of the body about the axis 
through the center of gravity parallel to the axis of sus- 
pension, then 

rr\-\ n -i rC -p CL 

I hereiore i = ; — , 

a 

or d^l -d)= k\ 

In this formula c?and I — d enter in exactly the same way. 
It follows therefore that if 0' were taken as a center of 
suspension, would be the center of oscillation. That 
is, in a compound pendulum the centers of oscillation and 
suspension are interchangeable. This is easily verified ex- 
perimentally. 

Problem 423. In a plane through the center of gravity, (7, of a 

body circles of radii r and — are drawn with centers at C, where k is 

r 

the radius of gyration about a gravity axis perpendicular to the 
given plane. Show that the time of vibration is the same for all 
axes of suspension perpendicular to the given plane and passing 
through a point on the circumference of either circle. 

184. Experimental Determination of Moment of Inertia. — 

The computation of the moment of inertia of many bodies 
is a difficult matter. It is often convenient, therefore, to 
use an experimental method in dealing with such bodies. 
The compound pendulum furnishes a means whereby such 
determinations may be made. From Art. 182, we find 
that the time of vibration of a compound pendulum is 

^gd 



DYNAMICS OF RIGID BODIES 347 

This may be written 

IT 

Multiplying both sides by M^ the mass of the body, we 
have 

It thus appears that if d^ the distance from to the 
center of gravity, is known (the center of gravity may be 
located by balancing over a knife edge) and also the weight 
(7, and the body be allowed to swing as a pendulum about 
as an axis, t may be determined, giving Iq. 

If Ig be desired, it may be determined from the formula 
(see Art. 64), 

Problem 424. The connecting rod of a high-speed engine tapers 
regularly from the cross-head end to the crank-pin end. Its length is 
10 ft., its cross section at the large end 5.59" x 12.58" and at the 
cross-head end 5.59" x 8.39". Neglecting the holes at the ends, the 
center of gravity is 64 in. from the cross-head end. The rod is made 
of steel and vibrates as a pendulum about the cross-head end in 1.3 sec. 
Compute its moment of inertia about the gravity axis. 

The student should take such a connecting rod as the one in the 
preceding problem, or other body, and by swinging it as a pendulum 
find its period of vibration. Compute the moment of inertia about 
the axis of suspension and about the gravity axis. 

185. Determination of g. — From the preceding article 
we see that 

M«r2 77^2 

^ df MtH 



348 



APPLIED MECHANICS FOR ENGINEERS 



This relation enables us to determine g, as soon as we know 
Iq, M, and d, by determining the time of vibration about the 



point 0, 



It is evident that -^ — 



Md 



is a constant for the 



body, when the axis is through 0, and that when once 
determined accurately, the pendulum might be used to 
determine g for any locality. 



L 



TT 



This constant, -^-—s is known as the pendulum constant. 

Md 

Problem 425. A round rod of steel 6 ft. long is made to swing as 
a pendulum about an axis tangent to one end andperpendicular to its 
length. The rod is 1 in. in diameter. Determine the pendulum 
constant. 

Problem 426. The center of gravity of a connecting rod 5 ft. 
long is 3 ft. from the cross-head end. The rod is vibrated as a pen- 
dulum about the cross-head end. It is found that 50 vibrations are 
made in a minute. Find the radius of gyration with respect to the 
cross-head end. 



186. The Torsion Balance. — A torsion 
balance consists of a body suspended 
by a slender rod or wire attached 
rigidly to the body and at the point of 
support, the center of gravity of the 
body lying in the line of the wire (Fig. 
262). 

Let OA be the neutral position of a 
line in the surface of the disk used as 
a torsion pendulum (Fig. 262). 

Let the disk be turned through an 
angle 6^ and released. The wire is 




Fig. 262 



DYNAMICS OF RIGID BODIES 349 

then twisted and exerts a twisting moment on the body, 
tending to restore it to the neutral position. Experiment 
shows that this twisting moment exerted by the wire is 
proportional to the angle of twist. Hence, since 

2 mom = /a, 

we have for the motion of the disk, starting from the posi- 
tion where 6 = ^q, 

-oo = i<^f,, (1) 

the minus sign being used since co -— is the acceleration 

du 

counter-clockwise and CO represents the clockwise twist- 
ing moment. 

.'. Icod(o=- OOde. 
Integrating, 

2 2 ^ 

When t=0,e=eQ,co = 0. .-. (7i=C^. 

.-. Ja)2= a(6'2-6'2). " (2) 

This equation shows that the angular velocity of the 
body is the same for negative values of 6 as for the nu- 
merically equal positive values. The motion is therefore 
periodic, the body vibrating through equal angles on both 
sides of the neutral position. 

Considering the motion from ^ = ^^ to ^= 0, equation (2) 
may be written 

de W 






V^2-6I2 ^I 



350 APPLIED MECHANICS FOR ENGINEERS 

Integrating, 

0- \-j-^-^' 



sin-i- = - x/- • ^ + a. 





When t=0, = 0^. r. C^ = sin-i 1. 

Choosing — as the value ofsin~il, we have 

This equation shows that as t increases sin"i— - must de- 

crease. Hence, since we chose sin~^"- = — when 6 = 6^^^ 

6q 2 

the angle — must decrease from — and will therefore reach 
zero when ^ = 0. 

/y 

Hence t = -^\yy when ^ = 0. 

Therefore the time for the body to make a complete swing, 
one way, is 



If m-^ is the twisting moment exerted by the wire when 
twisted through an angle 6^ m^ = C6^, and the expression 
for the period becomes 






The time of vibration is independent of the initial angu- 
lar displacement. 



DYNAMICS OF RIGID BODIES 



351 



187. Determination of the Moment of Inertia of a Body by 
Means of the Torsion Balance. — The moment of inertia of 
a body may be found by suspending it by a wire and ob- 
serving the time of vibration when used as a torsion 



m 



balance. The constant (7= -^ is a constant of the wire or 

rod and depends upon the material and diameter. Know- 
ing this constant, it would only be necessary to determine 
the period of vibration in order to find /. 

For practical purposes, however, it is desirable to elimi- 



nate from consideration the value 



m^ 

^ 



For this purpose 



suppose the disk provided with a sus- 
pended platform rigidly attached as 
shown in cross section in Fig. 263. Let 
t be its time of vibration and / its mo- 
ment of inertia about the axis of sus- 
pension. Now place on the disk two 
equal cylinders H in such a way that 
their center of gravity is the axis of 
suspension. Let ^j be the period of 
vibration of the cylinders and support 
and ij their moment of inertia. 

Then — =— . The moment of inertia of the two cyl- 
inders with respect to the axis of rotation is known ; 




Fig. 263 



call it ^. 


Then 


Ii = I+Iv 


so that 




I-I '" 



352 APPLIED MECHANICS FOR ENGINEERS 

This gives the moment of inertia of the torsion balance, 
which, of course, is a constant. 

The moment of inertia of any body L may now be 
determined by placing the body on the suspended plat- 
form with its center of gravity in the axis of rotation and 
noting the time of vibration. Calling the time of vibra- 
tion of the body L and the balance ^3 and their moment 
of inertia ig, we have 

t = L 
Let the moment of inertia of L itself be JT^, so that 

Then h = r-^- 

This method may be used in finding the moment of 
inertia of non-homogeneous bodies, provided the center 
of gravity be placed in the axis of rotation. 

Problem 427. The moment of inertia of a torsion balance is 
6300, where the units of mass, space, and time are respectively the 
gram, centimeter, and second, and its time of vibration 20 sec. The 
body L consists of a homogeneous cast-iron disk 3 in. in diameter 
and 1 in. thick. Find the time of vibration of the balance when L is 
in place. Compute the moment of inertia of the disk. 

Problem 428. The same balance as that used in the preceding 
problem is loaded with a body Z, and the time of vibration is found 
to be 30 sec. Determine the moment of inertia of L. 

188. Rotating Body under the Action of no Forces. — If a 

body is rotating about a fixed axis, the axis of 2, and is 
acted upon by no forces, the equations of Art. 180 become 



DYNAMICS OF RIGID BODIES 353 

= -(o^xM-a'^M, (1) 

= - (o'^M+ axM, (2) 

= 0, (3) 

= 0)2 CyzdM - a CxzdM, (4) 

() = -(o^C xzd M- a CyzdM, (5) 

0=aJ,. (6) 

From these equations it follows that 

a = 0, CxzdM=0, CyzdM =^0, x = 0, y = 0. 

Hence, if a body is rotating about a fixed axis and no 
forces are acting on the body, 

(1) the angular velocity is constant, 

(2) the axis of rotation passes throiigh the center of 
gravity, 

(3) if the axis of rotation is chosen as the axis of z, then 

CxzdM= 0, and CyzdM= 0. 

(When ixzdM=0 and I yzdM=0, the 2-axis is a prin- 
cipal axis of the body. It can be shown that there are 
three lines at right angles to each other through any point 
of the body which are principal axes, and that the ellipsoid 
of inertia of the body for tllat point has its axes lying on 
these three lines. Compare Art. 77.) 

189. Rotation of a Body about a Fixed Axis with Constant 
Angular Velocity. — If the angular velocity is constant, 
a = 0, and the equations of Art. 180 take simpler forms. 
2a 



354 APPLIED MECHANICS FOR ENGINEERS 

In addition, at a» given instant, let the a^-axis be chosen 
through the center of gravity of the body. Then ^ = 0, 
and equations (1) to (6) of Art. 180 become 

2X, = -G,2^i!f, (1) 

2 F, = 0, (2) 

2Z, = 0, (3) 

S inoni^^ = ft)2 I yzdM^ (4) 

2 momjj/ = — ft)2 J xzdM^ (5) 

2 mom,., = 0. (6) 

The first equation shows that unless x is zero there is a 
resultant of all the x-components of the impressed forces 
and that this resultant is 

R^=- uP'xM. 

Equations (2) and (3) show that the ^-components of the 
impressed forces either have a resultant zero or else form 
a couple, and the same of the ^-components. 
A particular case is that where 

i^xzdM=^ 0, and (yzdM= 0, 

as, for example, where the a^^-plane is a plane of symmetry. 
Suppose in this case the only impressed forces are the 
weights and the reactions of the supports, as indicated in 
Fig. 264. 

Equations (1) to (6) then become 

P^4-P; = -a)2^i^f, 0') 

Py^P[=^^. (2') 

P.-Tf=0, (30 



DYNAMICS OF RIGID BODIES 



355 



hPy - aP[ = 0, 

aP',^xW-hP, = 0, 

0=0. 




Fig. 264 



From (20 and (4'), Py =0, P^ = 0. 
From (3'), P, = TT. 



(4') 
(5') 
(6') 



From (!') and (5'), Px = 



a: 



a 






"" -: W ^—^xaP'M. 



p; = _ 



If the weight W were counterbalanced by an equal 
upward force acting through (7, the only forces acting on 
the axis would be 

P, = ^ xay'^M, PI = ^—- xco^M. 

a -{- a + 



356 



APPLIED MECHANICS FOB ENGINEERS 



The resultant of these forces is a single force, 

and if its distance above the a;^-plane is z\ 

z'R. = aPL-hP^ 



a -\-b a -\-h 



= 0. 



Therefore, when the a??/-plane is a plane of symmetry, or 
when j xzdM= and j yzdM = 0, the effect of the rota- 
tion of the body about the axis of z is to cause an outward 
pull on the axis of rotation in a line through the center 
of gravity perpendicular to the axis of rotation and of the 
same value, MxuP'^ that would be caused by concentrating 
the whole mass at the center of gravity. 

Problem 429. A wheel of weight 100 lb. is mounted on an axle 
and is off center \ inch, the plane of the wheel being perpendicular to 
the axis. Find the force tending to bend the shaft when the wheelis 
making 200 r. p m. 

Problem 430. A thin rod, 2 ft. long, weighing 5 lb. is attached to 
an axle at an angle of 60°. Find the outward 
force in magnitude and position in the plane of 
the rod and axle due to the rotation when mak- 
ing 150 r. p. m. If the rod is joined to the axle 
2 ft. and 1 ft. from the supports (Fig. 265), find 
the reaction at the supports due to the rotation. 

Problem 431. A steel disk 3 ft. in diameter 
and 1 in. thick is not perpendicular to the axis 
of rotation, but is out of true by -^ of its 
radius. Find the twisting couple introduced 
tending to make the shaft wobble. 




DYNAMICS OF RIGID BODIES 



357 



Problem 432. A grindstone weighing 200 lb., of radius 2 ft., is 
making 100 r. p. m. Find the force with which one half of the stone 
pulls on the other half. 

Problem 433. Neglecting the weight and 
the tension in the spokes of a rotating fly- 
wheel, prove that the tension in the rim is 



P = 



iip-r^ 



yF 



9 




Fig. 266 



where w = angular velocity, y = the heaviness 
of the material, F = the area of the cross 
section of the rim, and r = mean radius of 
the rim (Fig. 266). 

Problem 434. In the preceding problem, 
suppose r = 6 ft., F=10 in. by 4 in., and the 
wheel is made of cast iron. If the tensile strength of the material is 
25,000 lb. per square inch, what speed would be attained before the 
wheel bursts ? 

Problem 435. Show that if P is the pressure of the side rod of a 

r> locomotive on the 
crank pin, r the ra- 
dius of the crank-pin 
circle, r' the radius 
of the driver, and v 
the velocity of the 
train, 

Grv^ 




dN 



□cLY 




G 
Fig. 267 



2P-G^ 



/2 



gr 

when the side rod is 



in the lowest position (Fig. 267). 



Problem 436. Suppose the velocity of a locomotive to be 90 
mi. per hour, the radius of the crank-pin circle 20 in., the radius 
of the drive wheel 40 in., and the weight of the side rod 400 lb. 
Find the pressure on the crank pins due to the rotation alone. 



358 APPLIED MECHANICS FOR ENGINEERS 

190. Rotation of a Locomotive Drive Wheel. — The drive 
wheel of a locomotive (Fig. 268) may be considered for 
the present as rotating about a fixed axis. We shall con- 
sider the effect of the weight of the counterbalance on the 

tire due to rotation only, on the 
assumption that the tire carries all 
the weight of the counterbalance. 

Note. It is to be understood that the 
wheel center carries part of the weight of 
the counterbalance, but a complete solu- 
tion of the problem of the drive wheel is 
beyond the scope of this book. The above 
assumption is therefore made. 

Let M be the mass of J of tire 
and p the distance of its center of 
iG. 268 gj^.^^^^^ from the center of wheel. 

Let iHfj be the mass of the counterbalance, and p^ the dis- 
tance of its center of gravity from the center of wheel. 

Then 2 P = 0)2 (Mp + M^p^^. 

In particular, suppose the diameter of the tread of the tire 
to be 80 in., distance of the center of gravity of ^ of tire 
from center 27 in., and mass of ^ of tire 21. The mass 
of the counterbalance is 20, and the distance of its center 
of gravity from the center of the wheel 29 in. Substitut- 
ing these values, we get 

2P = a)2 [21 (f I) + 20 (II)] =95.6 6)2. 

If now we know the speed of rotation of the wheel so 
that CO is known, we may determine P. Let us take co 
corresponding to a speed of train of 60 mi. per hour. 




DYNAMICS OF BIGID BODIES 359 

This gives o) = 26.4 radians per second and 

P = 33,300 lb. 

Problem 437. If the area of the cross section of the tire is 20 
sq. in., find the stress on the metal due to rotation about the axis 
under the above assumption. 

If the allowable stress on the metal is 20.000 lb. per square inch, 
find the speed of train to cause this stress. 

191. Standing and Running Balance of a Shaft. — Let 
weights, TFj^ ^^ W^, W^, etc., be attached to a shaft, the 
distances of their centers from the center of the shaft 
being r^, r^, ^3, r^, etc. (Fig. 269), it being assumed that 




Fig. 269 

each weight is symmetric with respect to a plane through 
its center and perpendicular to the axis of the shaft, so 
that the pull of the weight on the shaft due to rotation is 
through the center of gravity of the weight (Art. 189). 

If the shaft is balanced when at rest in any horizontal 
position, the sum of the moments of the weights about 
the axis must be zero for any position. The moment of 



360 APPLIED MECHANICS FOR ENGINEERS 

Wi is Wi times the horizontal component of r^, or it is 
the horizontal component of W^r^ laid off along r^ The 
weights will then be in standing balance when, and only 
when, the sum of the horizontal components of the vec- 
tors r^TFj, r^W^, ^^3^1 etc., is zero for any horizontal posi- 
tion of the shaft. Hence the vector polygon 

must close. 

For running balance the reactions at each support, due 
to the centrifugal pull of the weights, must annul or the 
shaft will tend to wobble. These forces due to rotation 
are along the radii (Art. 189), and if x-^, x^, 2^3, ... are the 
distances along the shaft of the weights from one bearing, 

the reactions at the other bearing are -^r^ccP'M-^^ -^r^ctP'M^^ 

• .., where I is the distance between the supports, and M^, 
iHigi ^31 etc., are the masses of the weights. 

These reactions are therefore proportional to x^^W-^^ 
^2^2 ^2' ^^^"> ^^^^ ^^6 in the directions of r-^, rg, etc., re- 
spectively. In order to balance, the polygon of vectors 

^i^i^i + 'V2^2+--* 
must therefore close. 

In order that the reactions due to rotation balance at 
the other bearing, the vector polygon 

(I - x^y^ TFi -f (Z - x^y^ TF2 + . . . 

must close. But since the vectors may be taken in any 
order, this polygon may be plotted in the order 



DYNAMICS OF RIGID BODIES 361 

which clearly will close when the two polygons already 
considered close. 

There will then be standing and running balance when 
two closed polygons can be formed of vectors r^ TTj, r^ "B^, 
r^W^, etc., and x-^r^W-^, x^r^W^, x^r^W^^ etc., respectively, 
the vectors having the directions of the perpendiculars 
from the center line of the axle to the centers of gravity 
of the corresponding weights. 

Problem 438. In Fig. 269 assume W^ = 6, W^ = 5, TFg = 4 ; 
Tj = 4.3, rg = 3, Tg = 6, r^ = 2 ; angle between r^ and r^ = 135°, angle 
between r^ and r^ = 90° ; x^ = 1.5, 0:2 = 1. Find x^ and the value and 
position of W^ for running balance. 

Problem 439. Show that two weights, given in position and 
magnitude in the same plane perpendicular to the axis of rotation, 
can be balanced for running by a third given weight in that plane. 
Show how to determine the position of this third weight, and show 
that the condition for standing balance is the same as for running 
balance in the case of weights all in the same plane perpendicular 
to the axis of rotation. 

Problem 440. If two weights are attached to the axis in the same 
plane containing the axis, show how to find the position of a third 
given weight that will balance the two given weights for running and 
standing balance. If balanced for standing, will they be balanced for 
running ? 

Problem 441. Show that three weights cannot be balanced for 
running unless they either lie all in one plane perpendicular to the 
axis of rotation or else all in a plane containing the axis. 

Problem 442. Show that four weights and the arms of three of 
them being given and two of the weights definitely located on the axis, 
the other weights may be placed so as to form standing and running 
balance, and show how the location of the two weights would be de- 
termined. Could this be done in more than one way? 



362 



APPLIED MECHANICS FOR ENGINEERS 



192. Rotation of Flywheel of Steam Engine. — In Fig. 270 
let a belt run horizontally over a flywheel, the tensions 
being F^ and P^ where P^ > P^ The effective steam 
pressure is P, J^' the pressure of the guides on the cross- 
head. It is normal if friction is neglected. 




.— TTf^^EEE , -, H ^ni-^, 



r^ 



-2a- 



FiG. 270 



The pressure on the crank pin is resolved into tan- 
gential and radial components, T and i^j. 
From the relation, 

2 moms = aZ, 

we have for the motion of the wheel 

Ta - (P2 - ^1)^ = «^- (1) 

It will be assumed that the resistance of the machinery, 
as shown by P^ — Pj, is constant. 

When T=(^P^ — Pi)-, « = 0, and the angular velocity, 

CO, is either a maximum or a minimum. 

When ^=0 (Fig. 270), T=0, a is negative, and co is 



DYNAMICS OF RIGID BODIES 363 

therefore decreasing. Hence co has a minimum value in 

the first quadrant at B^ where T= (Pg — ^i)~' 

From B^ the angular velocity increases as long as T 

remains greater than {P^ — P^-, In the second quad- 

rant as T decreases toward zero at the dead point A, there 

is a point A^ where jr= (Pg — P\)- ^^ which therefore g) 

has a maximum value. In the same way the points A^ 
and B^ in the third and fourth quadrants correspond to 
minimum and maximum values of o). 
Equation (1) may be written 

7a)c?a) = TadS - {P^ - P^rdO, (!') 

If s and s' are the arcs passed over in the crank-pin circle 
and the flywheel circle respectively, then 

adO = d% and rdd = c?s', 

and equation (!') becomes on integrating from an initial 
value ft)Q to any value «, 

\ iQ^ - 0,2) = frds - (P, - Pj) (s' - «;). 

Since the work done on one end of the connecting rod 
equals the work done by the other, neglecting its own 
change in kinetic energy, 

CTds = Cpdx, 

and hence 



364 APPLIED MECHANICS FOR ENGINEERS 

Pdx may be found by read- 

ing from the indicator card the values of P for successive 
values of x between the limits x and x^. 

A different treatment of the above equation may be 
obtained by considering that the pressure of steam in 
the cylinder is constant and equal to P' up to the point 
of cut-off and that beyond this point the pressure varies 
inversely as the volume. If we assume P constant and 
equal to P' to the cut-off, then the limits of integration 
will be regarded accordingly, and we may write 

J /(<»f - 0,2) = P'ix^ -x,-)-(F,^ Pi) (.s-i - «;>, 

where x-^ is the value of x at the cut-off. 

Beyond the cut-off P varies inversely as the volume of 
steam in the cylinder, or 

P = ^P'. 

X 

Then from the point of cut-off to any value of x, 

^f'l X 

= x,P' log, (^) - (P, - PO (s' - s[). 

If P[ is the mean effective pressure, we have, considering 
the work done on the flywheel for the motion from B to 
A^ neglecting friction and the change in kinetic energy of 
the connecting rod, 

1 7(0,2 - 0,1) = 2 aP[ - (P, - P,-)7rr. 

Problem 443. Suppose the mean effective steam pressure is 16,000 
lb., the radius of the crank -pin circle 18 in., and the radius of the fly- 



DYNAMICS OF RIGID BODIES 



365 



wheel 3 ft. If P^- P^= 500 lb., /"^ = 2000, and w^ = 2 tt radians per 
second, find co^. 

Problem 444. The flywheel in the above problem has a velocity 
(0^ = 6 TT radians per second. What constant resistance (Pg — -^i) "^ill 
change this to 2?? radians per second in 100 revolutions? 

Problem 445. Find the values of for which w is maximum and 
minimum in the above problem. 

Problem 446." In Problem 443 find the value of w when the wheel 
has turned through 90° from B. 

Problem 447. If the total pressure in the above problem is 20,000 
lb. up to the cut-off at } the stroke, find <o^. 

193. Rotation and Translation. — Let be a point in a 
body having plane motion, i.e. translation and rotation in 
a plane, and let the angular 
velocity and angular ac- 
celeration of the body be co 
and a respectively (Fig. 
271). 

At any instant choose 
axes with origin at and 
the a;-axis along the vector 
of <x, the linear acceleration 
of the point 0. Any element of the body, distant r from 
0, then has the acceleration a parallel to the a:-axis, a 
tangential acceleration ra, and a radial acceleration rco\ 
and the corresponding effective forces acting on dM are 
adM, radM, and rwHM. 

Applying D'Alembert's principle, taking the z-axis 
through perpendicular to the plane of motion, 

^Xi= C(adM- rco'^dM cos (l)-radM sin (f)}, 




Fig. 271 



366 



APPLIED MECHANICS FOR ENGINEERS 



2 r;- = C(radM cos - rco^dM sin <^), 

which reduce to 

^Xi = Ma- ayM- (oHM, 
^Yi = axM-(o'^yM, 
2 monifg = al^ — ayM. 



(1) 

(2) 
(3) 



194. The Connecting Rod. — The connecting rod of an 
engine has a circular motion at one end while the other 
end moves backward and forward in a straight line. We 
shall consider the motion relative to the engine and shall 
assume that the flywheel is of sufficient weight to give 
the crank a motion sensibly uniform. It will be con- 
venient to regard the motion of the connecting rod as con- 
sisting of a rotation about the crosshead end while that end 
is moving in a straight line. 

In Fig. 272 let A be the crosshead and the center of 
the crank-pin circle. Let I be the length of the connect- 
ing rod, and r the radius of the crank-pin circle. If we 
neglect friction, the only forces acting on the connecting 




DYNAMICS OF RIGID BODIES 367 

rod at A are iV', the pressure of the guides, and P', the 
pressure exerted by the piston rod. The force exerted on 
the connecting rod by the crank pin has been resolved 
into its normal and tangential components iV^ and T, re- 
spectively. Suppose ft)j to be the constant angular velocity 
of the crank, and suppose the angular velocity of the rod 
to be represented by (o and the angular acceleration by a. 
Let a be the linear acceleration of the crosshead. 

The equations of the preceding article applied to the 
connecting rod become 

F'- Tsm0-]}^^cosO = Ma-co^xM-a^M, (1) 

-N'- a-\-]^^sm 6- Tcos 6 ^-oy^yM^- axM, (2) 

N^l sin ((9 + <^) - Tl cos (<9 + ^) - i ai cos cj) 

= al-ayM. (3) 

Before these equations can be solved for the reactions, the 

values of «, a, and a must be known. These values can 

be expressed in terms of g)^. 

From the figure, 

r sin 0=1 sin 0. 

Differentiating, 

r cos u —- = I cos 9 -^ • 
at at 

But —- = (!)- ana-^=a). 

dt ^ dt 

. *. ra)j cos = Ico cos (^, 

or 0)= ^"ico8e_ ^4N 

V«2 - r.2 siii2 e 

Differentiating again, we obtain 

_d(a _ (P — r'^)r(d\ sin 



(5) 



368 APPLIED MECHANICS FOR ENGINEERS 

To obtain a notice that any two points of the rod have 
velocities whose components along the rod are equal. 
Hence if v-^ is the velocity of the crank pin and v is the 
velocity of the crosshead, 

V cos <^ = -^1 sin (d-\-(j)'). (Fig. 272) 

. •. V = v^ (sin 6 + cos 6 tan (^). 

dv ( c^dQ ' a X. J d^^ , z) 9j dd> 

a = —- = v.[ cos u — - — sin u tan 9 h cos u sec^(p —^ 

dt \ dt dt dt. 



= V 



"cos (^ + <f)) . cos 

^ ^1 "1 o ^ 

COS <!> COS^ (j) 



or a = -^^ [«i cos (e + <|>) cos <f) + « cos 6] . (6) 

cos'<j> 

The value of co from equation (4) may be substituted in 
equation (6) and thus the value of a determined. 

Problem 448. The connecting rod given in Problem 424, Art. 
184, is in use on an engine whose crank has a constant angular 
velocity of 26 radians per second. The length of the crank is 2 ft., 
the effective steani pressure on the piston is 16,000 lb. Use the values 
of M, I, X, and y from Problem 424. Find the values of N', N^, and T 
when e = 30°. 

Problem 449. Show that w has its greatest numerical value when 

o 

^ = and 9 = TT, and its least numerical value when 6 =— and 6 = — . 

2 2 

What are these greatest and least values? 

Problem 450. Find what values of 6 will make a a maximum or 
minimum. Locate the crosshead for these values. 

Problem 451. Find values for T, N', N-^, and the resultant pressure 
on the crank pin when — tt and when ^ = 0. Use the above data. 

Problem 452. Assume a force of friction F acting on the cross- 
head, such that F = .06 N'. In the above case when 6 = 30°, what is 
the value of F, N', N^, and T? 



DYNAMICS OF BIGID BODIES 



369 




Fig. 273 



Problem 453. Suppose the steam pressure zero, find T, N', N^, 
and the resultant crank-pin pressure, if w^ is the same. 

195. Angular Momentum or Moment of Momentum. — If 
the velocity of a particle is resolved into two components, 
one of which is parallel to a given 
line and the other in a plane perpen- 
dicular to the line, the product of the 
latter component, the perpendicular 
distance of this component from the 
given line, and the mass of the particle 
is called the moment of momentum or 
angular momentum of the particle with 
respect to tlie given line. Thus, if 
v^ is the component of the particle in 
the plane perpendicular to the given line, d the distance 
of this component from the given line, and m the mass of 
the particle, then its 

angular momentum = mv-^d. (Fig. 273) 

If the given line be taken as the axis of x^ the component 
Vj may be resolved into components Vy and v^ parallel to 
the y- and 2-axes, respectively. Then, since the moment 
of any vector is equal to the sum of the moments of its 
components, w^e have 

angular momentum of the particle with respect to 2:-axis 

= (yv^ — zvy)m. 

If dm is the element of mass of any body, then the angu- 
lar momentum of the body about any line is defined to be 
the sum of the angular momenta of its particles about that 
line. Thus, representing the angular momenta of a body 

2 B 



370 APPLIED MECHANICS FOR ENGINEERS 

about the x-^ 3/-, and 2-axes by hj,. hy, h^^ respectively, we 
have 

196. Torque and Angular Momentum. — Let dm be the 
mass of an element of a body, and a^., a^, a^, respectively, 
the components of its acceleration parallel to the axes. 

Using D'Alembert's principle, we may equate the sum 
of the moments of the effective forces to the sum of the 
moments of the impressed forces. 

The effective forces acting on dm are a^dm^ aydm^ a^dm^ 
respectively parallel to the x-^ y-^ and 2-axes. Hence call- 
ing the torques due to the impressed forces about the axes 
respectively T^., Ty^ T^, we have 

T^ = J {ya^ - zay^dm, 

Ty= I {za^ — xa^dm^ 

T^= \{xay — ya^dm. 
^-r dx dy dz 

Now ., = -, ^ = j, .. = -, 

_ ^'^x _ dVy _ dv^ 

^ dt ^ dt " dt 

Also -TXy'^z — ^'^y)='^y'"z + y^z—'^zVy — zay=ya^ — zay. 

'd 



'■• T^=jj^iy^z-zvy)dm. 



Since the derivative of a sum of terms is the sum of the 
derivatives of the separate terms, and the integral is the 



DYNAMICS OF RIGID BODIES 371 

limit of a summation, we may write this latter equation 



Likewise Ty = -^^, 



or rp^^dh, 

at 

d\ 
dt 

and T, = ^'' 

dt 

That is : The moment of impressed forces about any axis is 
equal to the rate of change of the angular momentum about 
that axis. 

A corollary of this theorem is that when the moment of 
impressed forces acting on a body about an axis is con- 
stantly equal to zero, the angular momentum about that 
axis is constant. 

197. Moment of Momentum of a Body with One Point 
Fixed in Terms of Angular Velocity. — If a body in motion 
has one point fixed, its motion at any instant is one of 
instantaneous rotation about an axis passing through the 
fixed point. Take the fixed point, 0, of the body as 
origin of coordinates. The angular velocity about the 
instantaneous axis can be resolved into component 
angular velocities, oj^., a)^, co^, about the axes (Art. 132). 

The velocity v^ or — is then due to rotation about OY 

^ dt 

and OZ. Figure 274 shows the component parallel to OX 
of the velocity due to rotation about OZ to be 

— r-^(Oz cos /3 or — yw^. 



o72 APPLIED MECHANICS FOE ENGINEERS 



In the same way the rc-component of the velocity due to 



rotation about OY is zcoy. 



Wz 



z 




dx 
di 



= ^^y - y^z 



Similarly 
and 



dy 

-&- = XQ), — 2a)_, 

dt 



dz 
dt 



y^X - ^03y. 



Substituting these values in the expressions for A^, we 
obtain 

— I (j/^ + z'^)o)^dm — 1 xywydm — i xzco^dm^ 
or Aj. = 1^03^ — Wyi xydm — (oA xzdm. 

Similarly, hy = IyOi>y — (nA yzdm — (o^l yxdrn^ 
and hz = I^(o^ — (^x\ ^xdm — (Oyi zydm. 



DYNAMICS OF RIGID BODIES 373 

The only body whose motion we shall study in this book 
is a uniform body in the form of a solid of revolution. If 
one of the axes be taken as the axis of the body, then all 
of the products of inertia vanish and the expressions for 
the angular momenta become 

198. Vector Representation of Angular Momentum. — Sim- 
ilar to the representation of angular velocity by a vector, 
angular momentum about any line may be represented by 
a vector along the line. Since in the special case we shall 
study the angular momentum about any axis used is pro- 
portional to the angular velocity about that axis, it is clear 
that the angular momentum may properly be treated as 
a vector in the way defined. The signs of the angular 
momenta will be the same as those of the corresponding 
angular velocities, and the vectors will be laid off in the 
same way as for angular velocities. 

199. Kinetic Energy of a Body with One Fixed Point. — 
The kinetic energ}^ of a body at any instant is the sum of 
the kinetic energy of its particles, or 

K. E. = ^ Cv^dm = 1 C^v^ +vl-{- vl^dm. 
In Art. 197 it was proved that 

The substitution of these values in the expression for 
kinetic energy gives 

K. E. = ljl(f + /)a,J + (a;2 ^ ^2^)^2 ^_ ^^2 _^ x'^)(o';]dm 



374 APPLIED MECHANICS FOR ENGINEERS 



J [zi/Q)yCo^ H- xzoo^o)^ + xi/co^Qyy^dm, 



or, K. E. = I J o>2 + 1 J 0)2 + ^ J a>2 



— Wy«g j zydm — (>>^o>z i xzdtn, — "jpO^ j ocydm. 




\ 



For the special case of the body of revolution with one 
of the coordinate axes coinciding with the axis of the 
body, or for any body when referred to principal axes, 

K. E. = - J «2 + i J «2 + - J 0)2 

^* ^* 2 ^x X ^ 2 y y ^ 2 ^z z- 

200. Vector Rate of Change Due to Rotation. — Let h be 

a vector with point of application at 0. Suppose the 

vector is changing with the time 
in magnitude and direction but 
keeps the point of application the 
^^' same. Let the vector change in 

length from li to h-\- Ah in time A^ and in that time turn 

through the angle A^. The rate of change in the original 

direction of the vector is then 

Limit (^ "^ A^)cos AO — h 

^t^ Alt 

and the rate of change perpendicular to the original direc- 
tion of the vector is 

A/=0 A^ 

The first of these quantities may be written 



Limit 



-A(l-cosAg) M^^^^^- 
At At 



DYNAMICS OF RIGID BODIES 



375 



= Lim 



= Lim 



h '2 ' sin^ 



-h 



A0- 



At 



+ 



dh 
~di 



. Ad 

sm — 

2 . AS AO 
sm — • — 

AS 2 A^ 



+ 



dh 



dt 
_ dh 
~ dt' 



dh 
dt 



The second is 



A^^o^ ^ A^ A^ 



= A • 1 • — 
= Aft), 

where g) is the angular velocity of rotation of the vector. 
Hence the rate of change in the direction of the vector is 

dh 
di' 

and the rate of change at right angles to the vector is 

hco. 

201. Rate of Change of Angular Momentum of a Body Due 
to Rotating Axes. — Let the components of the angular mo- 
mentum of a body with fixed point with respect to axes 
OX, OY^ OZ be respectively Aj, Ag^ ^31 ^^^ ^^^ ^^^^ frame 
of axes coincide at a given instant with a fixed frame OXp 



37() 



APPLIED MECHANICS FOR ENGINEERS 



0T\, OZ^^ but be rotating about the fixed frame with 

angular velocities tWj, co^^ co^. 

Lay off the vectors 7ij, Ag^ ^h ^^^ ^^^ moving axes (Fig. 

276). Since the moving axes at the given instant coincide 

with the fixed axes, the com- 
ponents of the momentum are 
the same for the fixed and 
moving axes at that instant. 
The rate of change of angular 
momentum along the fixed 
^ axes is then given by the 
method of the preceding 
article. Thus along OX^ the 



z^ 



h. 



X, 



Y 



rate of change is — l, due to 



Fig. 276 



the change in h^ This is in- 
creased by Ag&)2 by the rotation of Ag about the i/-axis and 
by — ^2^3 by the rotation of h^ about the 2-axis. Thus the 
rate of change of angular momentum about the fixed x- 
axis is 



dt 



- h^(Oo + AoQ)o. 



2rr 



Similarly about the fixed ?/-axis the rate of change is 

dK 



dt 



^ — AgCOj + 7^lCt> 



3' 



and about the fixed s-axis the rate of change is 

dt ^ ^ ^ ^ 



DYNAMICS OF RIGID BODIES 



377 



Since the moment of impressed forces about any axis is 
equal to tiie rate of change of the angular momentum about 
that axis, we may write 

dhc 



.2«^3 



.3^,2, 



^.= -^- Vl+^«1«3' 



T, 



dJi^ 
dt 



h^a)^ + /igCOj. 



dm. 



ax 



202. Motion of the Center of Gravity of Any Body. — Let 

dm be the mass of any element of a body and (x^ y, 2) the 
coordinates of dm (Fig. 277). The 
effective forces acting on dm are re- 
spectively 

a^dm, aydm^ and a/Lm, 

Equating the components of the im- 
pressed and effective forces acting on 
the body, parallel to the a:;-axis, we 

have 

2Xi = ^ajim. (1) 

Now if X is the abscissa of the 
center of gravity of the bod}^ 

Mx — ^xdm. 
Differentiating, 

j^dx ^ ^dx^ 
dt dt 




Fig. 277 



and 



(i"X dj X 

M — - = S — dm = ^ajim. 
dt^ dt^ 



378 



APPLIED MECHANICS FOR ENGINEERS 



Therefore, if ^^ is the a;-component of the acceleration of 
the center of gravity of the body, equation (1) may be 
written 

2X, = Ma,. 

In like manner, 

■ ^Y, = Ma,, 
2Z,. = Ma,. 

Hence, for any motion of a body the acceleration of the 
center of gravity is the same as if the whole mass were con- 
centrated there and all the impressed forces were applied 
there parallel to their original directions. 

203. Gyroscope. Simplest Case — Consider a body in the 
form of a solid of revolution turning with uniform angular 
velocity, o), about its geometric axis, which is horizontal, 

while at the same time 
^ this axis, passing always 

through a fixed point, 
turns with uniform an- 
gular velocity, II, about 
the vertical (Fig. 278). 
The problem is to de- 
termine the forces neces- 
sary to maintain this 
motion. 

Let 00 he the axis of the body, moving with the body. 
Choose fixed axes OX, OY, OZ, so that the a;-axis 
coincides with 0(7 at the given instant. 

Let OA be the horizontal axis perpendicular to 00. 
At the given instant OA coincides with OY. 




Fig. 278 



DYNAMICS OF RIGID BODIES 379 

Denote the moment of inertia of the body about OChj (7, 
and that about OA or OZ by A. Then from Art. 197 the 
angular momenta about 00^ OA^ and OZ slvq respectively 

h^ = (7ft), ^2 =0, Ag = AD^. 

These values of A^, h^^ and Ag, while they are the values of 
the angular momenta about the fixed axes OX^ OY, OZ, 
at the given instant, are not values which hold in general, 
and cannot therefore be differentiated to obtain the torques 
about these axes. The method of Art. 201 applies here 
where 



0)^ =0, (^2= 0, ft>g = 12. 




We may write then 




^x= ^^^ Agftjg 4-^30)2 = 0, 


(1) 


Ty = — -^ — AgWj H- Ajft>g = (7ft)ll, 


(2) 


y, = ^ - ^i<»2 + ^2»i = 0- 


(3) 



For the motion of the center of gravity of the body, since 
the center of gravity is moving uniformly in a horizontal 
circle and is at the given instant on the ic-axis, the acceler- 
ation is directed toward the point of support along the 
rc-axis. Hence 

a^. = — bD,\ ay = 0, a^= 0, 

where h is the distance from the point of support to the 
center of gravity. 

.-. 2X, = -itfm2, (4) 

2:F, = 0, (5) 

2Zi=i). (Art. 202) (6) 



380 APPLIED MECHANICS FOR ENGINEERS 

The impressed forces necessary to satisfy equations (1) 
to (6) may, under certain conditions, be only the weight 
of the body and the reaction of the support. If these are 
the only impressed forces, equation (2) becomes 

Wb = (7a>n, 

where W is the weight of the rotating body. The re- 
maining equations are then satisfied, so that the necessary 
and sufficient condition that the body have the motion 
described is 

C(aQ = Wb. 

The motion of the body around the vertical axis is 
colled, precession. When the angular velocity around the 
vertical axis is constant, the precession is said to be steady. 

Problem 454. If the rotating body is a thin disk 2 ft. in diam- 
eter fastened to a rod of negligible weight which passes through the 
center of the disk perpendicular to its faces, the center of the disk 
being 6 in. from the point of support, what will be the rate of pre- 
cession in a horizontal plane when the angular velocity of rotation 
of the disk about its axis is 300 r. p. m. ? Ans. 9.79 r. p. m. 

Problem 455. How would the rate of precession vary with the 
radius of gyration of the disk if the angular velocity co be kept un- 
changed ? 

204. The Gyroscopic Couple. — Equation (2), Art. 203 
shows that when a body of revolution is rotating uniformly 
about its axis and this axis is rotating uniforml}^ about 
an axis at right angles to it, there is acting on tlie body 
at right angles to the other two axes a couple whose value 
is the product of the moment of inertia of the body about 
its axis of revolution and the two angular velocities. This 



DYNAMICS OF RIGID BODIES 381 

couple is called the gyroscopic couple. In the case of the 
body discussed in Art. 203, the gyroscopic couple is (7a)Il. 
Since action and reaction are equal and opposite, the 
rotating body will resist the turning of its plane of rota- 
tion with a couple equal and opposite to the gyroscopic 
couple. This finds application in the effect of rapidly 
rotating wheels of cars, automobiles, etc., when turning 
around curves. In Fig. 278 if the disk represents a wheel 
of a car rolling in the direction indicated by co and at the 
same time turning about OZ., the rotation about OZ would 
be in the opposite direction; i.e. H would be negative 
in the formula Ty = CcoH. The gyroscopic couple acting 
on the wheel would then be in the opposite direction, or 
about 01^ in the direction from OX to OZ. The couple 
that the wheel would exert, tending to upset the car, would 
then be about OYiw the direction from OZ to OX. 

Problem 456. A ship carries a cast-iron flywheel whose rim is 
6 ft. outside diameter, 4 in. thick, and 18 in. wide. When it is 
making 3 revolutions per second, its axis is turned about an axis 
through the plane of the wheel with unit angular velocity. Find the 
moment of the couple that tends to turn the wheel about an axis 
perpendicular to these two axes. 

Problem 457. A solid cast-iron disk 3 ft. in diameter and 3 in. 
thick revolves about its axis, making 3000 revolutions per minute. 
At the same time it is made to turn about an axis in its plane at 
the rate of 2 revolutions per minute. Find the magnitude of the 
couple tending to rotate the disk about an axis perpendicular to these 
two axes. 

Problem 458. A locomotive is going at the rate of 40 mi. per 
hour around a curve of 600 ft. radius. The diameter of the drivers 
is 80 in., and a pair of drivers and axle have a moment of inertia 



382 APPLIED MECHANICS FOR ENGINEERS 

about an axis midway between the wheels and perpendicular to the 
axle of 3000. What is the magnitude of the couple introduced by the 
precession al motion of this pair of wlieels ? Give the direction in which 
it acts. Does it tend to make the locomotive tip inward or outward ? 

Problem 459. A car pulled by the locomotive in the preceding 
problem has four pairs of wheels. The moment of inertia of each 
pair of wheels and their connecting axle, with respect to an axis mid- 
way between the wheels and perpendicular to the axle, is 320. 
What is the magnitude of the precessional couple acting upon tlie 
whole car? 

Problem 460. The flywheel of an engine on board a ship makes 
300 revolutions per minute. The rim has the following dimensions : 
outside radius 4 ft., inside radius 3| ft., width 12 in. The ship rolls 
with an angular velocity of ^ a radian per second ; find the toi'que 
acting on the ship due to the gyrostatic action of the flywheel. 

205. Car on Single Rail. — An interesting application 
of the gyroscope has been made recently in England. A 
car is run upon a single rail, and is held upright by means 
of rapidly rotating flywheels. Each car contains two of 
these wheels rotating in opposite directions, at the rate of 
8000 revolutions per minute. 

Any tendency of the car to tip over, either when running 
or standing at the station, is righted by the gyroscopic 
action of the flywheels. The experimental car was so 
successful in operation that it maintained itself in an 
upright position even when loaded eccentrically. The 
action of the flywheels is such as to place the center of 
gravity of the car and load directly over the rails. See 
Engineering^ June 7, 1907. 

Another practical application of the gyroscopic couple 
is to be found in Schlick's ''stabilizer" for steadying ships. 



DYNAMICS OF RIGID BODIES 



383 



206. Gyroscope. Inclined Axis. — Let the angular veloc- 
ity of the body relative to its own axis, 0(7, be « at any 
instant, the angular velocity with which the plane con- 
taining the axis turns about the vertical be 11, and the 
angle which the axis of the body makes with the vertical 
be 6. The component about 00 oi the total angular 
velocity of the body is then co + the component of 11 
about 00 or co 4- II cos 0. 

The angular velocity co is called the velocity of spin. 



Let 



dt 



= ft)^ 



Then (o' is the angular velocity of the 



axis of the body in the vertical plane containing the axis. 

Choose fixed axes so that 
at the instant in question 
the a:3-plane contains the 
axis of the body, the x- 
axis being horizontal (Fig. 
279). Let OA, OB, 00 he 
a set of moving axes, of 
which 00 is the axis of the 
rotating body, OB is hori- 
zontal, and OA is at right 
angles to 0(7 and OB. Let 
0^1, OB^, 00^ be fixed 
axes which at the given instant coincide with OA, OB, 
00 respectively. The frame of moving axes OABO then 
has angular velocities about the fixed axes 0^4^, OB^, 00-^ 
respectively 

ft)j = — fl sin 6, 0)^ = co', cOq = fl cos 0. 

The angular velocity 12 may be resolved into components 




384 APPLIED MECHANICS FOR ENGINEERS 

about OA^ OB, (9 C' respectively 

— XI ain ^, 0, O cos 0. 

The angular velocities of the body about the axes OA, 
OB, O' respectively are therefore 

— fl sin 0, w\ and co + O cos 6. 

If the moments of inertia of the body about the axes 
OA, OB, 00 are respectively A., B, C, or, because of sym- 
metry, A, A, 0, the angular momenta of the body about 
these axes are respectively 

A-^ = — A^ sin 6, h^ = Aco' , h^ = 0(^co + n cos 0). 

The discussion will be divided into two cases : (a) steady 
precession, (5) unsteady precession. 

(a) Steady Precession. Suppose the angle and the 
angular velocities co and XI are all constant. For this case 
ft)' = and 

\ = — AD. sin 6, h^ = 0, Ag =: (7(ft> + XI cos ^), 

J = — XI sin 6, 0)2 = 0, cog = XI cos 6. 



CO 



Also, equating the torques and the rates of change of the 
angular momenta about the axes OA^, OB^, OC-^, we have 

Ta = ^ - VS + h'->2 = 0' (1) 

at 



(O 



^^ ^ lit ~ ^^"^^ ^ ^^"^ 

= (7X1 sin 6'(ft> + XI cos 6) - AD'^ sm d cos 6, (2) 



DYNAMICS OF RIGID BODIES 385 

It is possible for this motion to take place under the 
action of the weight and the reaction of the support as 
the only impressed forces for certain related values of 
ft), fl, and the constants of the body. For if only the 
weight and the support act on the body, then, since the 
center of gravity moves as if all the forces were concen- 
trated there, the only additional conditions to be satisfied 

are 

X, = -Mb sin (902, (4) 

r; = o, (5) 

Zr = TF, (6) 

where ^^, T^, Zj. are the components of the reaction of the 
support along the fixed axes at the given instant, and b is 
the distance from the point of support to the center of 
gravity of the body. 

With only the weight and the reactions of the support 
as impressed forces, the torques T^ and T^ are both zero 
and Tf, = Wb sin 6, All of the conditions of motion would 
then be satisfied if, from equation (2), 

Wb sin 6 = on sin 6(co + H cos 6) - An"^ sin cos (9, 

or {A - 0) cos en'^ - Ccoa -^ wb = o. 

The values of H that satisfy this equation are 



^^ Cft)±VaW-4 Wb^A- (7)cos6> 
2{A- (7) cos l9 

There are therefore two solutions, one solution, or no 
solution according as 



C^co^^^WbCA- (7)cos6>. 



2c 



386 APPLIED MECHANICS FOR ENGINEERS 

Problem 461. Find the value, or values, of O for steady pre- 
cession with axis inclined 75° from the vertical of the disk of Prob- 
lem 454, the thickness of the disk being negligible. 

What is the least value of o) that would satisfy the conditions for 
the motion of this disk when = 75°? 

Problem 462. A conical top is made of wood and is spinning 
about its axis with a velocity of 20 revolutions per second. The cone 
has a base of 2 in. and a height of 2 in., and spins on the apex. 
While spinning steadily with its axis vertical (sleeping), it is dis- 
turbed by a blow so that its axis is inclined at an angle of 30' with 
the vertical. Find the velocity of precession and the torque that tends 
to keep the top from falling. 

(J) Unsteady Precession. Assume 0, co^ and O to be 
variables. The equations then become 

a)j = — 12 sin 0, co^ = co', cOg = II cos 0, 

h^ = ~ An sin ^, ^2 = Act)', h^ = C(^(o + U, cos ^), 

T^ = - Afn COS 0'(o' -\- sin e—\-Aco'n cos 6 

^ ^^ ^ -^Cco' (co +n cos 0), (1') 

Tj, = A^-\- en sin 0(co + n COS 6) - An^ sin cos 0, (2^ 

r, = (7^(0) + O cos 6>). (3') 

etc 

Again suppose the weight and the reaction of the 
support to be the only impressed forces. Then T^ — 0, 
Ti, = Wb sin 6, T^ = 0. Equation (3^ then becomes 

4-(«4-Xlcos^)=0. 
at 

Therefore 

(o -\~ n cos ^ = a constant. 



DYNAMICS OF RIGID BODIES 387 

For the sake of simplicity suppose the body to have 
been placed with its geometric axis at rest making an 
angle 6^ with the vertical, the body then given an angular 
velocity coq about its axis and released. Then the value of 
ft) + fl cos 6 at the start was g)q. Therefore throughout 
the motion 

ft) + H cos 6 = (Oq. 

Substituting this value in equation (1^) where ^^ = 0, 
and replacing co' by — , it becomes 

(a/V 

2 An cos 6dd + A sin edfl - Cco^dd = 0. 

If this equation be multiplied through by sin 6^ it can then 
be integrated, for it becomes 

^(2 X2 sin e cos Odd + sin2 Od^^ - 0(o^ sin SdO^ 0, 

in which the quantity inclosed in the parenthesis is 
^(O sin2 (9). 
Integrating, 

An sin^ 6 + Ocdq cos 6 = constant. 

At the beginning of motion = 0, and 6 = 6^. 
Therefore 

An sin^ 6 = (7a)Q (cos 6^ — cos ^). (5') 

This equation expresses H in terms of 6. 

Instead of using equation (2') a simpler equation is ob- 
tained by considerations of work and energy. The work 
done from the beginning of motion to the instant under 
consideration is that done by gravity, i.e. 

TTKcos^o- cos l9). 



388 APPLIED MECHANICS FOR ENGINEERS 

The gain in kinetic energy in this time is 

\iA(- n sin (9)2 + ^A(o'^ 4- 1- CcoH - i Ccol (Art. 199) 

Therefore 

lA(n^ sin2 e + a)'2) = Tf 6 (cos 0^ ~ cos (9), 

2Wb 
or ft)'2= (cos ^0 — cos ^) — 112 sin2 ^^ 

A 

Replacing cos 6^ — cos by — -— from equation (5'), 

Ccoq 

co'^ = n(^ -n) sin2 6. (6') 

\ CcOq J 

The factors 12 and — H must both be of the same sign 

since 0)^2 u^ust be positive. Assuming &)q to be positive, it 
follows that 12 is positive, for if XI were negative, the sec- 
ond factor would be positive and their product would be 
negative. 

Also the values of 12 that make to' zero are and 

From equation (5') 

O — ^^ cos 6^ — cos Q 
~ A ' sin2 6> 

from which it can be shown that 



C(o^ 



'A— 
dd A sin3 d 



[(cos6'o-cosl9)2 + sin2(9o]. 



Since sin 6 is positive, —zr is positive; i.e. 12 increases as 

du 

6 increases and decreases as decreases. It follows then 
that 6 and 12 both increase until 12 = • (o' is then 0, 6 



DYNAMICS OF BIGID BODIES 389 

changes from increasing to decreasing and continues to 
decrease until 11 = and = 6q (Equations (6') and 

(5'))- 

The axis of the body therefore alternately falls and rises 

between the values 0=6^ and 6 = 6^, where ^^ is the value 
obtained by putting Q. = —- — in equation (5'). As the 

axis of the body falls and rises, the value of H increases 

from when ^ = ^q to — — when 6 = ^j, decreasing again 

Ccoq 

to when 6 = 6^. 

The condition for this motion is that the value of 6 ob- 
tained from equations (5') shall be real. 

Problem 463. Using the body of Problem 454, making wq = 600 
r. p. m. and Oq = 30°, find the range of values for and O. 

Problem 464. For what least value of o)q would the body of the 
preceding problem have the motion described in this article, the other 
conditions being the same ? 



CHAPTER XV 

IMPACT 

207. Definitions. — When two bodies collide, they are said 
to be subject to impact. 

When the velocity of the striking surfaces is in the direc- 
tion of the normal to those surfaces, the impact is said to be 
direct. Otherwise it is oblique. 

When the normal to the surfaces at the point of contact 
(or center of forces exerted by one body on the other) 
passes through both centers of gravity of the bodies, the 
impact is said to be central. 

The phenomena of impact may be best studied by con- 
sidering the two bodies somewhat elastic. Suppose for 
simplicity that they are two spheres, M-^ and M^ (Eig. 280), 
and that they are moving with velocities v^ and v^ and 
that the impact is central. In Fig. 280 (a) they are shown 
at the instant when contact first takes place, and in Fig. 280 
(5) they are shown some time after first contact when each 
has been deformed somewhat by the pressure of the other. 
The dotted lines indicate the original spherical form and 
the full lines the assumed form of the deformed spheres. 
When the spheres first touch, the pressure P between them 
is zero, but as each one compresses the other, the pressure 
P increases until it becomes a maximum. The compression 
of the spheres is indicated in the figure by d-^ and d^. We 

390 



IMPACT 



391 




(a) 



shall designate the time during which the bodies are being 
compressed as the period of deformation. 

After the compression has reached its maximum value 
the bodies, if they be partially elastic, begin to separate and 
to regain their original 
form. The common ^ 

pressure P decreases 
and becomes zero, if the 
bodies are sufficiently 
elastic so that they 
finally separate. We 
shall designate this pe- 
riod of separation as the 
period of restitution^ and 
the velocities after sep- 
aration as vl and v^. 

If the bodies are en- 
tirely inelastic, there will 
be no restitution. They 
will, in that case, remain 
in contact just as they 
are when the pressure be- 
tween them is amaximum 
and will move on with a 
common velocity V. 

In what follows velocities and accelerations toward tlie 
right will be called positive and those toward the left 
negative. 

208. Direct Central Impact. — When the bodies meet in 
direct central impact, separation will take place along the 




(6) 
Fig. 280 



392 APPLIED MECHANICS FOR ENGINEERS 

line joining the centers of gravity. Let T be the time 
from the first contact up to the time of maximum 
pressure, that is, the time of deformation^ and T^ the time 
from first contact up to the time of separation. Then 
T^— T represents the time of restitution. We have, from 
Art. 103, dv = a ' dt. Considering the motion of M^ 
during the period of deformation, we have dv = adt and 

\ dv = — —— I Pdt. 

or M^iV- v^) = - Cpdt, 

where Vi^ the common velocity of the centers of gravity 
of the bodies at the end of the period of deformation. 
In a similar way, 

^2(^-^2)= r^dt, 

Pdt on the right-hand side of the 



preceding equations cannot be determined since we do not 
know in general how the pressure P varies with the time ; 
we do know, however, that they are equal term for term, 
so that we may eliminate them. We have, then, 

or (ilTi + iHf 2) V = MxVi + MiVi. 

If the impact is not too severe, elastic or partially elastic 
bodies tend to regain their original shape after the defor- 
mation has reached a maximum and finally separate if they 



IMPACT 393 

possess sufficient elasticity. Using the notation of Art. 
207, we have, for the period of restitution, if It is the force 
of restitution, 



M^j 'dv=:j Rdt, 



so that ^i(^i- ^) = - r ^Rdi^ 

and ^2(4 - F) = r ^Rdt, 

from which {M^ + M^ V = M^v\ + M^v^^. 

The value of the integral I Pdt during deformation will 

not in general be the same as its value during restitution. 

Call the ratio of I Hdt to I Pdt^ e. This value, which 

is called the coefficient of restitution^ is constant for given 
materials. It is unity for perfectly elastic substances, zero 
for non-elastic substances, and some intermediate value 
for the imperfectly elastic materials with which the en- 
gineer is usually concerned. The following values of e 
have been determined : for steel on steel, e = .bb \ for cast 
iron on cast iron, g = 1, nearly ; for wood, e = 0, nearly. 

From the above definition of e, it is seen at once that we 
may write 





-V 

1 
-^1 


and e ■■ 


V- 


V 


n which it follows 


that 








v[ = 


:r(i 


+ 0- 


-ev^. 





394 APPLIED MECHANICS FOR ENGINEERS 

where r=^;t±^^. 

From the above equations it is seen that 

209. Momentum and Kinetic Energy in Impact. — When a 
constant force acts upon a body for a certain time, the 
change in the velocity of the body is directly proportional 
to the force and to the time and inversely proportional to 
the mass of the body. 

The product of the force, F^ and the time, ^, is called 
the impulse of the force^ and the product of the mass and 
velocity is called the momentum of the body. (The body 
is supposed to have no rotation.) 

From Newton's law, F = Ma, it follows that 

Ft = M(v^ - ^;o), 

or, when a body is acted upon hy a constant force, the im- 
pulse of the force is equal to the change in momentum of the 
body. 

If the force is not constant, and is represented at any 

time by P, then | Pdt is the impulse of the force for 

the time from to ^, and from the equations of the preced- 
ing article. 



.r 



Pdt=-M^(iV-v^^, 



rPdt = M^{^V-v^\ 



IMPACT 395 

it follows that the impulse of the force is equal to the 
change of the momentum of the body on which it acts, no 
other force being assumed to act on the body. 
From the equations of the preceding article, 

(i^i + M^^ V= M^v^ + M^v^, 
and (ilffj + ilfg) ^= ^A + ^^2' 

it follows that there is no change in the sum of the mo- 
menta of the bodies in impact, either during the period 
of deformation or the period of restitution. This is 
otherwise evident from the fact that the forces acting on 
the bodies are equal and opposite, and hence what one 
body gains in momentum the other loses. 

With the kinetic energy it is different. During the 
period of deformation the loss in kinetic energy is 



_ 1 

~~ 2 



11-^ 22 M^ + M^ 



= ^^i^^2 ^, _ , ^2 

2(ilfi + 7^2) 

During the period of restitution there is a gain in kinetic 

_ e^M^M^ ^ _ .2 

~2(Jf, + i/2) ' ""' 
since v[ — v'^= — e(^v^ — v^). 

The total loss in kinetic energy during the impact is there- 



396 APPLIED MECHANICS FOR ENGINEERS 

If the bodies are inelastic, e = 0, and the bodies go on 
with the common velocity V. The loss in kinetic energy 
in this case is 

If the bodies are perfectly elastic, g = 1, and the loss in 

kinetic energy is zero. 

Problem 465. A lead sphere whose radius is 2 in. strikes a large 
mass of cast iron after falling freely from rest through a distance 
of 100 ft. What is its final velocity ? What is the loss of kinetic 
energy ? Assume e = 0. 

Problem 466. A 10-lb. lead sphere is at rest when it is acted 
upon by another lead sphere, whose radius is 3 in., in direct central 
impact. The velocity of the latter sphere is 20 ft. per second. What 
is the common velocity of the two spheres and what is the loss of 
kinetic energy due to impact ? 

Problem 467. Prove that if two inelastic bodies, moving in 
opposite directions with speeds inversely proportional to their masses, 
collide, both will be reduced to rest. 

Problem 468. Prove that if two perfectly elastic bodies with 
equal masses collide, each mass will have after impact the velocity 
that the other had before. If one of the bodies is at rest and the 
other strikes it, what will happen ? 

Problem 469. Figure 281 represents a set of perfectly elastic 

*- balls of equal mass and size in contact 

^ ■ ^ sA A A A A /) jjj a straight line. Another ball, of 

the same mass and material, moving 
in the direction of the line of balls, 
strikes one end of the line. Prove 



— ^-^^^ N A A A AA ) that the moving ball will be brought 

^^^- ^^^ to rest where it strikes and the last 

ball at the other end of the line will take its velocity, all other balls 
remaining at rest. Prove also that if two balls, moving together. 



IMPACT 397 

strike the line, the two balls at the other end of the line will take the 
velocity of the striking balls and the latter will be brought to rest. 

Problem 470. Prove that if a ball of mass Mi fall from a height 
H upon a very large mass with a horizontal surface and rebounds to 
a height h, the coefficient of restitution of the materials used is 



=V 



H 



(Regard M2 as infinite and V = 0.) 

Problem 471. A bullet weighing 1 oz. is fired horizontally into a 
box of sand (inelastic) weighing 20 lb., and remains imbedded in the 
sand. The box is suspended by a string attached to a fixed point 
4 ft. from the center of the box of sand. The impact of the bullet 
causes the box to swing aside through an angle of 42". Find (a) the 
velocity of imj^act of the bullet, {h) the kinetic energy lost in impact, 
(c) the greatest tension in the string. 

Problem 472. Assuming inelastic impact, show that if a pile 
driver weighing G-^ lb. falls h ft. and drives a pile weighing G^ lb. s ft. 
into the ground against a constant resistance R, 



R = 



_ G\ 



Gi+ G2S 



(Hint. Equate work done against resistance plus kinetic energy 
lost in impact to the work done by gravity on the falling weight.) 

Problem 473. A wooden pile weighing 1500 lb. is driven by a 
steel hammer, of weight 2000 lb., falling 20 ft. The penetration at 
the last blow is observed to be \ in. What is the resistance offered 
by the earth to the pile ? With a safety factor 6, what load would 
the pile carry ? 

210. Elasticity of Material. — All materials of engineer- 
ing are imperfectly elastic. Some, however, show almost 
perfect elasticity for stresses that are rather low. This 
has been expressed by saying that all materials have a 



398 APPLIED MECHANICS FOR ENGINEERS 

limit (elastic limit) beyond which if the stress be increased 
the material will be imperfectly elastic. Within the 
limit of elasticity^ stress is proportional to the deforma- 
tion produced. Let I^ be the total stress in tension or 
compression, / the stress, in pounds per square inch of 
cross section, d be the deformation caused by P and X 
the deformation per inch of length. Within the limit 

of elasticity of the material the ratio - is a constant, and 

P d ^ 

since / = — -, and \ = -, when F is the area of cross sec- 

tion and I is the length of the material, it may be written 

PI . . 

— — . This constant is called the modulus of elasticity of 

the material; it is usually represented by U, so that 

X Fd 

for tension or compression. For steel U has been found 
to be about 30,000,000 lb. per square inch. 

211. Impact Tension and Impact Compression. — Figure 
282 (a) represents a mass iltfg subjected to impact from 
the mass M^ falling from rest through a height h. The 
mass M^ is compressed by the impact. Figure 282 (6) 
represents the body M2 as subjected to impact in tension, 
the mass in this case being a rod having M-^ attached to one 
end and the other end attached to a crosshead A. The 
rod, crosshead, and weight fall freely together through 
the distance h until A strikes the stops at B^ when one end 
of the rod suddenly comes to rest and the mass M^ causes 
tension in the rod due to impact. 



IMPACT 



399 



Suppose v^ represents tlie velocity of M^ when impact 
occurs and l^ the length of M^^, whether it be a tension or 
compression piece. 



M, 




B 



2Io 



J/„ 






(a) 



(ft) 



Fig. 282 



In case (5) when the weights strike the stops at B the 
kinetic energy of the falling weights is equal to 

((^i4-^2)^ft.-lb., 

where (r^ and G-^ are the weights of M^ and i^ in pounds 
and h is in feet. This kinetic energy is used up in doing 
work in compressing the stops and in stretching M-^ and 
M^. If M^ is a rod of small cross section, its length is in- 
creased a large amount compared to the change in length 
of M^ and the stops against which the crosshead strikes, 



400 APPLIED MECHANICS FOB ENGINEERS 

and we may without serious error assume that all the ki- 
netic energy is used up in stretching M^. We may assume 
too that the stress in M^ is the same throughout its length. 

Let P^ be the maximum stress induced in the rod. The 

p 

average stress is then — ^, and if c?^ is the total increase in 

the length of the rod, the work done on the rod is 

2 
P (J 



m^-^m 



2 



=(a^ + a^)h. 



But p^^F^^%, (Art. 210) 

• • CC'-TM — ' ~: — — — • 



Here E must be expressed in pounds per square foot and 
all dimensions in feet if G^ and G-^ are in pounds and h in 
feet. 

In case (a) the student can easily show that 



U"m — 



"V 



2 Gihl 



F2E2 



Problem 474. Derive the formula just given for the case (a). 

Problem 475. A weight of 500 lb. falls through a distance of 2 ft. 
in such a way as to put a 1-in. round steel rod in tension. If the rod 
is 18 ft. long, what will be the elongation due to the impact? Use 
E = 30,000,000 lb. per square inch for steel. What maximum unit 
stress is induced in the rod ? 



IMPACT 401 

Problem 476. A cylindrical piece of steel 1 in. high and 1 in. in 
diameter is subjected to compression by a weight of 20 lb. falling 
through a distance of 1 in. How much will it be compressed? 

Problem 477. In the preceding problem what stress (pounds per 
square inch) was caused in the cylindrical block by the fall of the 
20-lb. weight? 

Problem 478. The safe stress in structural steel for moving loads 
is usually taken as 12,500 lb. per square inch. Through what height 
might a 300-lb. weight fall so as to produce tension in a 1-in. steel 
round bar, 10 ft. long, without exceeding the safe stress ? 

Problem 479. Two steel tension rods in a bridge, each 2 sq. in. in 
cross section and 20 ft. long, carry the effect of the impact of a loaded 
wagon as one wheel rolls over a stone 1 in. high. The weight on the 
wheel is 2000 lb. What stress is introduced in the tension rods ? 

I^OTE. For the strength of metals under impact the student is 
referred to the work of W. K. Hatt, Am. Soc, " Testing Materials," 
Vol. IV, p. 282. 

212. Direct Eccentric Impact. — Let M^ and M^ have 
impact as shown in Fig. 283, in which the centers of 
gravity of the bodies are moving along 
parallel lines and the surfaces of contact 
are perpendicular to the line of motion. ^ 
This is known as direct eccentric impact. ^ 

Let the velocity of M^ before impact 
be v^, the velocity of the center of gravity 
of M^ be Vj, and its angular velocity be 



ft) 



3/, 



1- 

(a) The problem will first be solved 

on the assumption that M^ is acted upon 

by no forces except that of impact. 

First period of impact. Let P be the force at any in- 
2d 



402 APPLIED MECHANICS FOR ENGINEERS 

stant of the first period of impact, T the time of the first 
period, F'the velocity of iUfgi (^ the angular velocity of M^^ 
and V^ the velocity of the center of gravity of M^ at the 
end of the first period. 

During the first period, which lasts for a very short 
time, the position of the body M^ may be regarded as 
unchanged. We then have for the motion of i^^, 

_ rPdt = M^ Cdv = M^( V- v^), (1) 

and for the motion of M^ 

CPdt = M^ C'dv = Jfi( Fj - Vi), (2) 

and hP = J^ 

dt 

or hrFdt = MJf:^C''d(o = M^k\(o-(o{), (3) 

where h is the radius of gyration of M^ about a gravity 
axis perpendicular to the plane of motion. Also, since at 
the end of the first period the velocity of M^ is the same 
as the velocity of the point of impact on M^^ 

■V=V^ + bco. . (4) 

The unknowns in these four equations are 






The equations are linear in these unknowns and the un- 
knowns can be easily determined from them. 

Second period ofimp,act. If R is the force of impact at 
any time during the second period of impact, T the value 
of ^, ^2 the velocity of M^^ v[ the velocity of the center of 



IMPACT 



403 



gravity of M^, and coj the angular velocity of M^ at the end 
of the second period, then 

-J^' Edt = M^J"' dv = M^{v^^ - F), (5) 

y Rdt = M^^^'^dv = M^{v[ - r^), (6) 

h^''^ Rdt = M^k^j^'' d(o = M^k\(o[ - CO) . (7) 

Rdt = el Pdt. (8) 

7^ «/o 



Also 



Here again are four linear equations in the four un- 

J<»Tf 
Rdt, vl^ v!), ojj, 
T 

which are therefore sufficient to determine the unknowns. 

(6) Impact on a body with fixed axis. Suppose the body 
rotating about the fixed axis through (Fig. 284), with 
angular acceleration a^ and angular 
velocity ©j at the beginning of im- 
pact. 

Choose the origin of coordinates 
at and the rr-axis coinciding with 
the direction of motion of the cen- 
ters of gravity of the bodies. 

Let X and Y be the reactions of 
the supporting axis through at 
any time during the impact. 

First period of impact. For the 
first period we have the following equations : 




Fig. 2»1 



404 APPLIED MECHANICS FOB ENGINEERS 

j^\P-hX)dt = M,(V,-v,\ (2') 

fJ[Pb - X(h - b)]dt = M^k\cD - a)i). (30 

Also, since is fixed, 

Fi = (A-5)a,, (4') 

V=h(o. (50 

Here are five equations containing five unknowns: 

f^Pdt, £'xdt, Fi, r, CO, 

from which the unknowns may be determined. 

Second period of compact. For the second period the 
following equations hold : 

- £' Bdt = M^iv', - V), (6') 

£\r + X)dt = M,(v[ - Fj), (7') 

C'lBb - X(h - *)]* = M-J<:\(o[ - o)), (8') 

v'i = (h-h)co[, (9') 

XT'/ /»r 

Edt = ej Pdt. (100 

Here again are five equations containing five unknowns : 
J Rdt, I Xdt, v[, v'^, (o[. 

These equations are sufficient to determine the unknowns, 
and thus the motion of the bodies is given after impact. 

In the above equations k is the radius of gyration of M^ 
about a gravity axis perpendicular to the plane of motion. 



IMPACT 405 

Problem 480. A uniform steel bar, 4 ft. long, weighing 16 lb., 
lies at rest on a smooth horizontal plane. It is struck by a steel ball 
weighing 2 lb., with a velocity of 30 f/s, at right angles to the rod at 
a point 6 in. from one end. Find the velocity of the ball, the velocity 
of the center of gravity of the bar, and its angular velocity at the end 
of the first period, and at the end of the second period of impact, 
given e = .55. 

Problem 481. In the preceding problem find the point about 
which the bar is turning at the end of the first period. Is this point 
the center of rotation at the end of the second period ? 

Problem 482. If Mi is at rest when M^ strikes with direct eccen- 
tric impact, write the equations which determine the motion at the 
end of the first, and at the end of the second period of impact. 

Problem 483. If e = between M^ and M^ of Art. 212 (a), find 

the kinetic energy lost in impact. 

Problem 484. Suppose Mx to be a rod of steel | in. in diameter 
and 2 ft. long, and suppose M2 to be a hammer weighing 2 lb. and 
that its velocity at the time of impact is 20 ft. per second, find V and 
<o, if the hammer strikes 10 in. from the center of the rod. e = .55. 

Problem 485. Let Mi be a square stick of timber 4" x 4" x 10' 
and let M2 be a 10-lb. hammer having a velocity at the time of impact 
of 10 ft. per second. If the impact takes place 4 ft. from the center, 
find V and w, given e = .10, 

213. Center of Percussion and Center of Instantaneous Rota- 
tion. — Consider a body at rest to be struck a blow and to 
be free to move in the plane passing through the center 
of gravity of the body and the line in which the blow is 
struck; as, for example, a slab at rest on a smooth hori- 
zontal plane when struck a blow in a horizontal plane 
through the center of gravity of the body. 

Let 6^ be the center of gravity of the body and P the 
force of the blow at any time during the impact (Fig. 



406 



APPLIED MECHANICS FOR ENGINEERS 



285). Then, since for the very brief period in which the 
blow takes place any change in the position of the body 

is negligible, we have for the 
motion of the body at any in- 
stant during the blow, if v is the 
velocity of the center of gravity, 
and CO the angular velocity, 

-dv 




P = M 



dt' 



dt 



Fig. 285 



where h is the radius of gyration 
about a gravity axis perpendicular to the plane of motion 
of the center of gravity of the body. 

If t is the time at any instant of the blow, measured 
from the beginning of the blow, there follow from the 
above equations, 



and 



By division, 



-i 



h\ Pdt=^ Mk^ 



'I d(o = 



Mk^< 



(O. 



V 

k^co 



The velocity of any point of the body at the given 
instant is composed of two components, one the velocity, 
V, of the center of gravity, and the other, rco^ relative to 
the center of gravity. There is then one point of the 
body, or in a plane fixed in the body and moving with it, 
where these two components just annul each other. This 
point (^, Fig. 285) must lie on the line through the 



IMPACT 407 

center of gravity perpendicular to the line of the blow at 
a distance a from the center of gravity such that 

a(o = V. 

By division, ab = k^, 

from which a is determined when h and k are known. 

There is, therefore, an axis for each blow about which 
the body begins to turn, and this axis remains approxi- 
mately the same as long as the change in the position of the 
body is negligible ; that is, in general, during the blow. 

The axis about which the body begins to turn under 
the action of the blow is called the instantaneous axis of 
rotation. 

The point, A^ in this axis which lies in the plane con- 
taining the center of gravity and the line of the blow is 
called the instantaneous center of rotation. 

If a fixed axis passed through the body at A., perpen- 
dicular to the plane containing the center of gravity and 
the line of the blow, the blow would cause the body to 
turn about this axis without causing any sudden reaction 
of the axis on the body. The point B., the foot of the per- 
pendicular from the center of gravity to the line of the 
blow, is known in this case as the center of percussion of 
the body corresponding to the center of rotation A. 

It should be noted that center of percussion and center 
of instantaneous rotation are related just as center of 
suspension and center of oscillation are in the compound 
pendulum (Art. 182). 



408 



APPLIED MECHANICS FOR ENGINEERS 



Problem 486. A thin rod, 3 ft. long, is struck a blow at 6 in. 
from one end and at right angles to the rod. Find the instantaneous 
center of rotation. 

Problem 487. A right circular steel cylinder of diameter 6 in. 
and height 2 ft. is suspended by an axis in a diameter of one end. 
It is struck a blow with a 5-lb. hammer with a velocity of 20 f/s 
through the center of percussion corresponding to the given support. 
Find the velocity of the hammer and the angular velocity of the 
cylinder at the time of greatest pressure and at the end of the impact, 
given e = .55. 

Problem 488. A right circular cone of steel, the radius of whose 
base is 6 in. and altitude 6 in., is supported as a pendulum by an axis 
through its vertex parallel to the base. It is struck with a 3-lb. 
hammer with a velocity of 10 ft. per second, in a line through the 
center of percussion. Find V and o> at time of greatest pressure. 

Problem 489. A man strikes a blow with a steel rod 1| in. in 
diameter and 4 ft. long, by holding the rod in the hand and striking 
the farther end against a stone in such a way as to cause the rod to 
be under flexure. Where should he grasp the rod in order that he 
may receive no shock ? 

Problem 490. Prove that after the blow has ceased to act on the 
body (Fig. 285) the body would then move in such a way that a circle 
of the body with radius CA and center C would roll along a straight 
line. Hence show that every point of the body in this circle would 
generate a cycloid. It is assumed that no other forces act on the body. 

214. Oblique Impact of Body against Smooth Plane. — Let 

M(Fig. 286) be a sphere 
moving toward the plane 
indicated with a velocity 
at impact of v, the direc- 
tion of motion making an 
angle a with the vertical 
to the plane. After im- 




IMPACT 409 

pact the body M rebounds with a velocity v^ in the direc- 
tion making an angle (3 with the vertical AB. Since 
the plane is considered smooth, the effect of the impact 
will be all in the direction of AB^ and hence the component 
of the velocity parallel to the plane will not change. 

. •. Vj sin l3 = V sin a. (1) 

Assuming that the plane does not move, F= 0, where V 
is the component of the velocity of M perpendicular to the 
plane at the time of greatest pressure. Then 



i 



^Pdt = M[0 - (y cos a)], 



ri 



and r Rdt = Mi- v^co^P -()'). 

Therefore 

J-*Tt 
Rdt 

vcosa rpdt 

or v-^ cos (3 = ev cos a. (2) 

Dividing (1) by (2), 

tan /8 = - tan a. 
e 

Squaring and adding (1) and (2), 

i»2 = ^2 (^sii;^2 a-{- e^ cos^ a) . 

Hence if a, v, and e are known, v-^ and ^ may be deter- 
mined. 

If the bodies are perfectly elastic, e = 1, /S = a, and 

v^ = V. If the body is inelastic, g = 0, /S = — , v^ = v sin a. 



410 



APPLIED MECHANICS FOR ENGINEERS 



The body then moves along the plane with a velocity 
V sin a. 

Problem 491. A ball is projected with velocity 50 f/s against a 
smooth plane surface at an angle with the normal of 40°. It leaves at an 
angle of 50° with the normal. Find e and the velocity at which it leaves. 

215. Impact of Rotating Bodies. — Suppose two bodies 
M^ and M2 revolve about two parallel axes Oj and O2 (Fig. 

287) in such a way 
that impact occurs at 
a point along the line 
DE. Let the line 
along which impact 
occurs be distant r^ 
and ^2 respectively 
from 0-^ and Og. Let 
P be the force of im- 
pact at any instant 
during the first period 
and M the force at any instant during the second period, 
ft) J and ft)2 the angular velocities at the beginning, co[ 
and CO2' the angular velocities at the end of impact, I^ 
and ig ^^^ moments of inertia of the bodies respectively 
about Oj and Og, and F'the rectangular component of the 
velocity of the point of impact in the direction of DJS at 
the time of greatest pressure. The angular velocities of 

the bodies at this instant are then — and — 




Fig. 287 



Then, as in Art. 212, 



(1) 



IMPACT 411 

'•iX^'^* = /iX"^<» = I,(a>[ - r^, (8) 

Rdt = e^^ Pdt. (5) 

From (1) and (2), 

4(F-ria,i)+:^^(r-r2a,2) = 0, 

or r= ^1^2(-^1^2^1+Vl^2) , 

From (1), (3), and (5), 
F fV 






or w[ = —(1 + g) — eft)!- 



Similarly, cog = —(1 + ^) — ^ft)2* 



(It should be noted that in these formulse the angular 
velocities of the bodies are reckoned in opposite direc- 
tions.) 

Problem 492. If I^= 3000, /g = 15,000, Wj = 1 radian per second, 
0)2 = 0, Tj = 2 ft., rg = 3 ft., and e = 0, find V, w{, Wg, and the kinetic 
energy lost in impact. 



412 



APPLIED MECHANICS FOR ENGINEERS 



Problem 493. A well drill is shown in principle in Fig. 288. 
The drill is supported by a cable that passes over a pulley C and is 
attached to a friction drum A. When A is held, the drill is raised 
by the operation of M^ and Afg. Suppose that / is 300 and cd^ = 3 




Fig. 288 



radians per second; /g = 200 and tOg^O; r^ = 2 ft. and r2 = 6 ft. 
Assume e = |. Find w[ and Wg. What kinetic energy is lost due to 
each impact ? 



IMPACT 



413 



Problem 494. The moment of inertia of the trip hammer M^, 
illustrated in principle in Fig. 289, is 100,000 ; that of ilfj is 60,000. 




Fig. 289 



If rj = 3 ft., r-i = 10 ft., (Oj = 2 radians per second, 0J2 = 0, and e = -|, 
find oj[ and Wo. What is the kinetic energy lost due to each impact? 
What is the kinetic energy of the hammer? 



APPENDIX I 

HYPEEBOLIC FUNCTIONS 



cosh X = — - — 

p^' p~ -^ 

sinh X = 

2 

, 1 sinh X p/' — e~^ 

tanh X = — = — 

cosh X e-^' -\- e ^ 



HYPERBOLIC FUNCTIONS 
1 



41' 



X 


Coshx 


Sinli X 


X 


Coslix 


Sinh X 


0.01 


1.0000500 


0.0100002 


0.51 


1.1328934 


0.5323978 


.02 


.0002000 


.0200013 


.52 


.1382741 


.5437536 


.03 


.0004500 


.0300045 


.53 


.1437686 


.5551637 


.04 


.0008000 


.0400107 


.54 


.1493776 


.5666292 


.05 


.0012503 


.0500208 


.55 


.1551014 


.5781516 


.06 


.0018006 


.0600360 


.56 


.1609408 


.5897317 


.07 


.0024510 


.0700572 


.57 


.1668962 


.6013708 


.08 


.0032017 


.0800854 


.58 


.1729685 


.6130701 


.09 


.0040527 


.0901215 


.59 


.1791579 


.6248305 


.10 


.0050042 


.1001668 


.60 


.1854652 


.6366536 


.11 


.0060561 


.1102220 


.61 


.1918912 


.6485402 


.12 


.0072086 


.1202882 


.62 


.1984363 


.6604917 


.13 


.0084618 


.1303664 


.63 


.2051013 


.6725093 


.14 


.0098161 


.1404578 


.64 


.2118867 


.6845942 


.15 


.0112711 


.1505631 


.65 


.2187933 


.6967475 


.16 


.0128274 


.1606835 


.66 


.2258219 


.7089704 


.17 


.0144849 


.1708200 


.67 


.2329730 


.7212643 


.18 


.0162438 


.1809735 


.68 


.2402474 


.7336303 


.19 


.0181044 


.1911452 


.69 


.2476458 


.7460697 


.20 


.0200668 


.2013360 


.70 


.2551690 


.7585837 


.21 


.0221311 


.2115469 


.71 


.2628178 


.7711735 


.22 


.0242977 


.2217790 


.72 


.2705927 


.7838405 


.23 


.0265668 


.2320333 


.73 


.2784948 


.7965858 


.24 


.0289384 


.2423107 


.74 


.2865248 


.8094107 


.25 


.0314132 


.2526122 


.75 


.2946833 


.8223167 


.26 


.0339908 


.2629393 


.76 


.3029713 


.8353049 


.27 


.03(56720 


.2732925 


.77 


.3113896 


.8483766 


.28 


.0394568 


.2836731 


.78 


.3199392 


.8615330 


.29 


.0423456 


.2940819 


.79 


.3286206 


.8747758 


.30 


.0453385 


.3045203 


.80 


.3374349 


.8881060 


.31 


.0484361 


.3149891 


.81 


.3463831 


.9015249 


.32 


.0516384 


.3254894 


.82 


.3554658 


.9150342 


.33 


.0549460 


.3360222 


.83 


.3646840 


.9286; 547 


.34 


.0583590 


.3465886 


.84 


.3740388 


.9423282 


.35 


.0618778 


.3571898 


.85 


.3835.309 


.9561 KiO 


.36 


.0655029 


.3678265 


.86 


,3931614 


.969S)ii!)3 


.37 


.0692345 


.3785001 


.87 


.4029312 


.9839796 


.38 


.0730730 


.3892116 


.88 


.4128413 


0.99.S0.-,S4 


.39 


.0770189 


.3999619 


.89 


.4228927 


1.0122369 


.40 


.0810724 


.4107523 


.90 


.4330864 


.0265167 


.41 


.0852341 


.4215838 


.91 


.4434234 


.0408^)91 


.42 


.0895042 


.4324574 


.92 


.4539048 


.0553S56 


.43 


.0938888 


.4433742 


.93 


.4645315 


.0699777 


.44 


.0983718 


.4543354 


.94 


.4753046 


.084(i7(i8 


.45 


.1029702 


.465.3420 


.95 


.4862254 


.0994843 


.46 


.1076788 


.4763952 


.96 


.4972947 


.1144018 


.47 


.1124983 


.4874<.I59 


.97 


.5085137 


.1294307 


.48 


.1174289 


.4<)S6455 


.98 


.5198837 


.1445726 


.49 


.1224712 


.5098450 


.99 


.5314057 


.159SL>,S8 


0.50 


1.12762()0 


0.5210953 


1.00 


1.5430806 


1.1752012 



418 



HYPEBBOLTC FUNCTIONS 
2 



X 


Oosh X 


Sinh X 


X 


Cosh X 


Sinh X 


1.01 


1.5549100 


l.lfX)6910 


1.51 


2.3738201 


2.1529104 


1:02 


.5668948 


.2062999 


1.52 


.3954676 


.1767566 


1.03 


.5790365 


.2220294 


1.53 


.4173563 


.2008206 


1.04 


.5913358 


.2378812 


1.54 


.4394857 


.2251046 


1.05 


.6037945 


.2538567 


1.55 


.4618591 


.2496111 


1.06 


.6164134 


.2699576 


1.56 


.4844787 


.2743426 


1.07 


.6291940 


.2861855 


1.57 


.5073467 


.2993014 


1.08 


.6421375 


.3025420 


1.58 


.5304654 


.3244903 


1.09 


.6552453 


.3190288 


1.59 


,5538373 


.3499117 


1.10 


.6685186 


.3356474 


1.60 


.5774645 


.3755679 


1.11 


.6819587 


.3523997 


1.61 


.6013494 


.4014618 


1.12 


.7005670 


.3642872 


1.62 


.6254945 


.4275958 


1.13 


.7093449 


.3863116 


1.63 


.6499021 


.4539726 


1.14 


.7232938 


.4034746 


1.64 


.6745748 


.4805947 


1.15 


.7374148 


.4207781 


1.65 


.6995149 


.5074650 


1.16 


.7517098 


.4382235 


1.66 


.7247249 


.5345859 


1.17 


.7661798 


.4558128 


1.67 


.7502074 


.5619603 


1.18 


.78082<)5 


.4735477 


1.68 


.7759650 


.5895910 


1.19 


.7956513 


.4914299 


1.69 


.8020001 


.6174806 


1.20 


.8106556 


.5094613 


1.70 


.8283154 


.6456319 


1.21 


.8258410 


.5276436 


1.71 


.8549136 


.6740479 


1.22 


.8412089 


.5459788 


1.72 


.8817974 


.7027311 


1.23 


.8567610 


.5644685 


1.73 


.9089692 


.7316847 


1.24 


.8724988 


.5831146 


1.74 


.9364319 


.7609115 


1.25 


.8884239 


.6019191 


1.75 


.9641884 


.7<K)4143 


1.26 


.9045378 


.6208837 


1.76 


2.9922411 


.8201962 


1.27 


.9208421 


.6400105 


1.77 


3.0205932 


.8502601 


1.28 


.9373385 


.6593012 


1.78 


.0492473 


.8806091 


1.29 


.9540287 


.6787578 


1.79 


.0782063 


.9112461 


1.30 


.9709143 


.6983824 


1.80 


.1074732 


.9421742 


1.31 


1.9879969 


.7181768 


1.81 


.1370508 


2.9733966 


1.32 


2.0052783 


.7381431 


1.82 


.16(>9421 


3.0049163 


1.33 


.0227603 


.7582830 


1.83 


.1971501 


.03673<i5 


1.34 


.0404446 


.7785989 


1.84 


.2276799 


.0688603 


1.35 


.0583329 


.799092() 


1.85 


.2585283 


.1012911 


1.36 


.07()4271 


.8197662 


1.86 


.2897047 


.1:340321 


1.37 


.0947288 


.8406219 


1.87 


.3212100 


.1670863 


1.38 


.1132401 


.8616615 


1.88 


.3530475 


.2004573 


1.39 


.1319627 


.8828874 


1.89 


.3852202 


.2341484 


1.40 


.1508985 


.9043015 


1.90 


.4177315 


.2681629 


1.41 


.1700494 


.9259060 


1.91 


.450584(1 


.3025041 


1.42 


.1894172 


■ .9477032 


1.92 


.4837827 


.3.371758 


1.43 


.2090041 


.9()9()951 


1.93 


.5173293 


.:3721810 


1.44 


.2288118 


1.9918840 


1.94 


.5512275 


.4075235 


1.45 


.2488424 


2.0142721 


1.95 


.5854808 


.4432067 


1.46 


.2690979 


.03(i8616 


1.96 


.6200927 


.4792343 


1.47 


.2895803 


.059()549 


1.97 


.6550667 


.5156097 


1.48 


.3102917 


.082(5540 


1.98 


. .61)04061 


.5523368 


1.49 


.3312341 


.1058614 


1.99 


.7261146 


.5894191 


1.50 


2.352409() 


2.1292794 


2.00 


3.7(J21957 


3.6268604 



HYPERBOLIC FUNCTIONS 



419 



X 


Cosh a; 


Sinh X 


X 


Coshcc 


Siuh X 


2.01 


3.7986528 


3.6646642 


2.51 


6.1930993 


6.1118311 


2.02 


.8354899 


.7028345 


2.52 


.2545281 


.1740685 


2.03 


.8727101 


.7413746 


2.53 


.3165827 


.2369237 


2.04 


.9103184 


.'7802896 


2.54 


.3792687 


.3004023 


2.05 


.9483548 


.8196198 


2.55 


.4425928 


.3645111 


2.06 


.9867111 


.8592571 


2.56 


.5065611 


.4292563 


2.07 


4.0255038 


.8993179 


2.57 


.5711800 


.4946444 


2.08 


.0647395 


.9398093 


2.58 


.6364560 


.5606820 


2.09 


.1043012 


.9806140 


2.59 


.7023958 


.6273758 


2.10 


.1443131 


4.0218567 


2.60 


.7690059 


.6947323 


2.11 


.1847398 


.0635018 


2.61 


.8362940 


.7627595 


2.12 


.2255846 


.1055530 


2.62 


.<)042644 


.8314615 


2.13 


.2668523 


.1480149 


2.63 


.9729254 


.9008469 


2.14 


.3085462 


.1908914 


2.64 


7.0422838 


.9709225 


2.15 


.3506713 


,2341871 


2.65 


.1123463 


7.0416950 


2.16 


.3932312 


.2779062 


2.66 


.1831184 


.1131701 


2.17 


.4362311 


.3220534 


2.67 


.2546108 


.185358(5 


2.18 


.4796741 


.3666325 


2.68 


.3268282 


.2582650 


2.19 


.5235649 


.4116482 


2.69 


.3997785 


.3318975 


2.20 


.5679083 


.4571052 


2.70 


.4734686 


.4062631 


2.21 


.6127086 


.5030079 


2.71 


.5479060 


.4813692 


2.22 


.6579702 


.5493610 


2.72 


.6230984 


.5572237 


2.23 


.7036972 


. .5961688 


2.73 


.6990531 


.6338338 


2.21 


.7498951 


.6434364 


2.74 


.7757775 


.7112072 


2.2o 


.79(35677 


.6911685 


2.75 


.853279!) 


.7893520 


2.26 


.8437197 


.7393692 


2.76 


.9315674 


.8682756 


2.27 


.8913565 


.7880444 


2.77 


8.0106482 


.9479862 


2^28 


.9394824 


.8371982 


2.78 


.0905297 


8.0284911 


2.29 


.9881022 


.8868358 


2.79 


.1712205 


.1097993 


2.30 


5.0372206 


.9369618 


2.80 


.2527285 


.1919185 


2.31 


.0868429 


.9875817 


2.81 


.3350617 


.2748566 


2.32 


.1369741 


5.0.387004 


2.82 


.4182283 


.3586224 


2.33 


.1876186 . 


.0903228 


2.83 


.5022368 


.4432239 


2.. 31 


.2387822 


.1424545 


2.84 


.587095(5 


.528(5699 


2.35 


.2905196 


.1951504 


2.85 


.6728130 


.(5149687 


2.36 


.;M26859 


.2482656 


2.86 


.759.3979 


.7021291 


2.37 


.3954365 


.3019558 


2.87 


.84(i8585 


.7901595 


2.38 


.4487266 


..3561760 


2.88 


.9352041 


.8790(594 


2.39 


.5025618 


.4109321 


2.89 


9.0241430 


.9(588(568 


2.40 


.5569472 


.4662293 


2.90 


.1145844 


9.0595611 


2.41 


.6118883 


.5220729 


2.91 


.2056373 


.1.511616 


2.42 


.6673910 


.5784683 


2.92 


.2976105 


.2436769 


2!43 


.7234594 


.6354226 


2.93 


.3905138 


.:!3711(58 


2.44 


.7801009 


.6929401 


2.94 


.484.3559 


.4314902 


2.45 


.8373201 


.7510265 


2.95 


.57914(37 


.52(58070 


2.46 


.89512.32 


.8096882 


2.96 


.6748952 


(32307(5.3 


2.47 


.95351.59 


.8689310 


2.97 


.7716115 


.720.3081 


2.48 


6.0125038 


.9287605 


2.98 


.86i):;047 


.8185119 


2.49 


.0720930 


.9891831 


2.99 


.9679850 


.917(597(5 


2.50 


6.1322895 


6.0502045 


3.00 


10.067(5620 


10.0178750 



420 



HYPEBBOLIC FUNCTIONS 



X 


Cosh X 


Sinlia; 


X 


Cosh X 


Sinh X 


3.01 


10.168.34.'56 


10.1190539 


3.51 


16.7390823 


16.7091854 


3.02 


.27004(54 


.2212451 


3.52 


,<K)70139 


.8774144 


3.03 


.3727741 


.3244585 


3.53 


17.07(56361 


17.0473312 


3.04 


.4765391 


.4287042 


3.54 


.2479662 


.2189529 


3.05 


.5813518 


.5339929 


3.55 


.4210213 


.3922966 


3.06 


.6872224 


.6403347 


3.56 


.5958178 


.5673790 


3.07 


.7941620 


.7477408 


3.57 


.7723744 


.7442186 


3.08 


.9021809 


.8562217 


3.58 


.9507082 


.9228325 


3.09 


11.0112900 


.9657881 


3.59 


18.1308371 


18.1032388 


3.10 


.1215004 


11.0764511 


3.60 


.3127790 


.2854552 


3.11 


.2328226 


.1882217 


3.61 


.4965523 


.4695004 


3.12 


.3452684 


.3011112 


3.62 


.6821753 


.6553927 


3.13 


.4588488 


.4151309 


3.f)3 


.8(596665 


.8431503 


3.14 


.5735748 


.5302919 


3.64 


19.0590447 


19.0327924 


3.15 


.6894584 


.(3466062 


3.(55 


.2503288 


.2243376 


3.16 


.8065107 


.7640850 


3.(36 


.4435377 


.4178052 


3.17 


.9247440 


.8827403 


3.67 


.6386909 


.6132145 


. 3.18 


12.0441695 


12.0025838 


3.68 


.8358083 


.8105854 


3.19 


.1647998 


.1236279 


3.69 


20.0349094 


20.0099373 


3.20 


.2866462 


.2458839 


3.70 


.2360140 


.2112905 


3.21 


.4097213 


.3693646 


3.71 


.4391421 


.4146645 


3.22 


.5340375 


.4940825 


3.72 


.6443142 


.6200802 


3.23 


.6596073 


.6200497 


3.73 


.8515505 


.8275577 


3.24 


.7864428 


.7472790 


3.74 


21.0608720 


21.0371178 


3.25 


.9145572 


.8757829 


3.75 


.2722997 


.2487819 


3.26 


13.0439629 


13.0055744 


3.76 


!4858548 


.4625710 


3.27 


.1746730 


.13(J6665 


3.77 


.7015584 


.6785064 


3.28 


.3067006 


.2690723 


3.78 


.9194324 


.8966096 


3.29 


.4400587 


.4028048 


3.79 


22.1394981 


22.1169025 


3.30 


.5747611 


.5378780 


3.80 


.3617777 


.3394069 


3.31 


.7108208 


.6743046 


3.81 


.5862933 


,5641452 


3.32 


.8482516 


.8120988 


3.82 


.8130681 


.7911403 


3.33 


.9870673 


.9512741 


3.83 


23.0421239 


23.0204143 


3.34 


14.1272820 


14.0918450 


3.84 


.2734843 


.2519907 


3.35 


.2689091 


.2338247 


3.85 


.50717J5 


.4858917 


3.36 


.4119630 


.3772277 


3.86 


.7432095 


.7221415 


3.37 


.5564583 


.5220(186 


3.87 


.9816222 


.9(507638 


3.38 


.7024094 


.()(i83619 


3.88 


24.2224327 


24.2017819 


3.39 


.8498306 


.8161219 


3.89 


.465(3658 


.4452205 


3.40 


.9987366 


.9(353634 


3.90 


.7113454 


.6911034 


3.41 


15.1491429 


15.1161016 


3.91 


.9594963 


.9394557 


3.42 


.3010637 


■.2683513 


3.92 


25.21C)1431 


25.15)03020 


3.43 


.4545147 


.4221278 


3.93 


.4(533109 


.4436673 


3.44 


.6095114 


.5774468 


3.94 


.7190247 


.6995765 


3.45 


.7660688 


.7:343232 


3.95 


.9773109 


.9580561 


3.46 


.9242033 


.8927735 


3.9(3 


26.2381943 


26.2191311 


3.47 


16.0839298 


16.0528128 


3.97 


.5017019 


.4828285 


3.48 


.2452646 


.2144571 


3.98 


.7(378597 


.7491740 


3.49 


.4082241 


.3777233 


3.99 


27.03(56943 


27.0181<)46 


3.50 


16..572H248 


16.542(5275 


4.00 


.3082331 


.289<)175 



APPENDIX II 

LOGARITHMS OF NUMBERS 



422 



LOGARITHMS OF NUMBERS 



LOGARITHMS OF NUMBERS, FROM TO 1000 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 








00000 


30103 


47712 


60206 


69897 


77815 


84510 


90309 


95424 


10 


00000 


00432 


00860 


01283 


01703 


02118 


02530 


02938 


03342 


03742 


11 


04139 


04532 


04921 


05307 


05690 


06069 


06445 


06818 


07188 


07554 


12 


07918 


08278 


08636 


089fK) 


09342 


09691 


10037 


10380 


10721 


11059 


13 


11394 


11727 


12057 


12385 


12710 


13033 


13353 


13672 


13987 


14)301 


14 


14613 


14921 


15228 


15533 


15836 


16136 


16435 


16731 


17026 


17318 


15 


17609 


17897 


18184 


18469 


18752 


19033 


19312 


19590 


19865 


20139 


16 


20412 


20682 


20951 


21218 


21484 


21748 


22010 


22271 


22530 


22788 


17 


23045 


23299 


23552 


23804 


24054 


24303 


24551 


24797 


25042 


25285 


18 


25527 


25767 


26007 


26245 


26481 


26717 


26951 


27184 


27415 


27646 


19 


27875 


28103 


28330 


28555 


28780 


29003 


29225 


29446 


29666 


29885 


20 


30103 


30319 


30535 


30749 


30963 


31175 


31386 


31597 


31806 


32014 


21 


32222 


32428 


32633 


32838 


33041 


33243 


33445 


33646 


33845 


34044 


22 


34242 


34439 


34635 


34830 


35024 


35218 


35410 


35602 


35793 


35983 


23 


36173 


36361 


36548 


36735 


36921 


37106 


37291 


37474 


37657 


37839 


24 


38021 


38201 


38381 


38560 


38739 


38916 


39093 


39269 


39445 


39619 


25 


39794 


39967 


40140 


40312 


40483 


40654 


40824 


40993 


41162 


41330 


26 


41497 


41664 


41830 


41995 


42160 


42324 


42488 


42651 


42813 


42975 


27 


43136 


43296 


43456 


43616 


43775 


43933 


44090 


44248 


44404 


44560 


28 


44716 


44870 


45024 


45178 


45331 


45484 


45636 


45788 


45939 


46089 


29 


46240 


46389 


46538 


46686 


46834 


46982 


47129 


47275 


47421 


47567 


30 


47712 


47856 


48000 


48144 


48287 


48430 


48572 


48713 


48855 


48995 


31 


49136 


49276 


49415 


49554 


49693 


49831 


49968 


50105 


50242 


50379 


32 


50515 


50650 


50785 


50920 


51054 


51188 


51321 


51454 


51587 


51719 


33 


51851 


51982 


52113 


52244 


52374 


52504 


52633 


52763 


52891 


53020 


34 


53148 


53275 


53402 


5352.t 


53655 


53781 


53907 


54033 


54157 


54282 


35 


54407 


54530 


54654 


54777 


54900 


55022 


55145 


55266 


55388 


55509 


36 


55630 


55750 


55870 


55990 


56110 


56229 


56348 


56466 


56584 


56702 


37 


56820 


56937 


57054 


57170 


57287 


57403 


57518 


57634 


57749 


57863 


38 


57978 


58092 


58206 


58319 


58433 


58546 


58658 


58771 


58883 


58995 


39 


59106 


59217 


59328 


59439 


59549 


59659 


59769 


59879 


59988 


60097 


40 


60206 


60314 


60422 


60530 


60638 


60745 


60852 


60959 


61066 


61172 


41 


61278 


61384 


61489 


61595 


61700 


61804 


61909 


62013 


62118 


62221 


42 


62325 


62428 


62531 


62634 


62736 


62838 


62941 


63042 


63144 


63245 


43 


63347 


63447 


63548 


63648 


63749 


63848 


63948 


64048 


64147 


64246 


44 


64345* 


64443 


64542 


64ti40 


64738 


64836 


64933 


65030 


65127 


65224 


45 


65321 


65417 


65513 


65609 


65705 


65801 


65896 


65991 


66086 


66181 


46 


66276 


66370 


66464 


-66558 


66651 


66745 


66838 


66931 


67024 


67117 


47 


67210 


67302 


67394 


67486 


67577 


67669 


67760 


67851 


67942 


68033 


48 


68124 


68214 


68304 


68394 


68484 


68574 


686()3 


68752 


68842 


68930 


49 


69020 


69108 


69196 


69284 


69372 


69460 


69548 


69635 


69722 


69810 


50 


69897 


69983 


70070 


70156 


70243 


70329 


70415 


70500 


70586 


70671 


51 


70757 


70842 


70927 


71011 


71096 


71180 


71265 


71349 


71433 


71516 


52 


71600 


71683 


71767 


71850 


71933 


72015 


72098 


72181 


72263 


72345 


53 


72428 


72509 


72591 


72672 


72754 


72835 


72916 


72997 


73078 


73158 


54 


73239 


73319 


73399 


73480 


73559 


73639 


73719 


73798 73878 


73957 



LOGARITHMS OF NUMBERS 



423 



LOGARITHMS OF NUMBERS, PROM TO lOOO 




(Cu7iti.nued) 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


55 


74036 


74115 


74193 


74272 


74351 


74429 


74507 


74585 


74663 


74741 


56 


74818 


74896 


74973 


75050 


75127 


75204 


75281 


75358 


75434 


75511 


57 


75587 


75663 


75739 


75815 


75891 


75966 


76042 


76117 


76192 


76267 


58 


76342 


76417 


76492 


76566 


76641 


76715 


76789 


76863 


76937 


77011 


59 


77085 


77158 


77232 


77305 


77378 


77451 


77524 


77597 


77670 


77742 


60 


77815 


77887 


77959 


78031 


78103 


78175 


78247 


78318 


78390 


78461 


61 


78533 


78604 


78675 


78746 


78816 


78887 


78958 


79028 


79098 


79169 


62 


79239 


79309 


79379 


79448 


79518 


79588 


79657 


79726 


79796 


79865 


63 


79934 


80002 


80071 


80140 


80208 


80277 


80;^5 


80413 


80482 


80550 


64 


80618 


80685 


80753 


80821 


80888 


80956 


81023 


81090 


81157 


81224 


65 


81291 


81358 


81424 


81491 


81557 


81624 


81690 


81756 


81822 


81888 


66 


81954 


82020 


82085 


82151 


82216 


82282 


82347 


82412 


82477 


82542 


67 


82607 


82672 


82736 


82801 


82866 


82930 


82994 


83058 


83123 


83187 


68 


83250 


83314 


83378 


83442 


83505 


83569 


83632 


83(-)95 


83758 


83821 


69 


83884 


83947 


84010 


84073 


84136 


84198 


84260 


84323 


84385 


84447 


70 


84509 


84571 


84633 


84695 


84757 


84818 


84880 


84941 


85003 


85064 


71 


85125 


85187 


85248 


85309 


85369 


85430 


85491 


85551 


85612 


85672 


72 


85733 


85793 


85853 


85913 


85973 


86033 


86093 


8()153 


8(i213 


86272 


73 


86332 


86391 


86451 


86510 


86569 


86628 


86687 


86746 


86805 


86864 


74 


86923 


86981 


87040 


87098 


87157 


87215 


87273 


87332 


87390 


87448 


75 


87506 


87564 


87621 


87679 


87737 


87794 


87852 


87909 


87966 


88024 


76 


88081 


88138 


88195 


88252 


88309 


88366 


88422 


88479 


8853(7 


88592 


77 


88649 


88705 


88761 


88818 


88874 


88930 


88986 


89042 


89098 


89153 


78 


89209 


89265 


89320 


89376 


89431 


89487 


89542 


89597 


89652 


89707 


79 


89762 


89817 


89872 


89927 


89982 


90036 


90091 


90145 


90200 


90254 


80 


90309 


90363 


90417 


90471 


90525 


90579 


90633 


90687 


90741 


90794 


81 


90848 


90902 


90955 


91009 


91062 


91115 


91169 


91222 


91275 


91328 


82 


91381 


91434 


91487 


91540 


91592 


91645 


91698 


91750 


91803 


91855 


83 


91!:K)7 


91960 


92012 


92064 


92116 


92168 


92220 


92272 


92324 


92376 


84 


92427 


92479 


92531 


92582 


92634 


92685 


92737 


92788 


92839 


92890 


85 


92941 


92993 


93044 


93095 


93146 


93196 


93247 


93298 


93348 


93399 


86 


93449 


93500 


93550 


93601 


93651 


93701 


93751 


93802 


93852 


93902 


87 


93951 


94001 


94051 


94101 


94151 


94200 


94250 


94300 


94M9 


94398 


88 


94448 


94497 


94546 


94596 


94()45 


94694 


94743 


94792 


94841 


94890 


89 


94939 


94987 


95036 


95085 


95133 


95182 


95230 


95279 


95327 


95376 


90 


95424 


95472 


95520 


95568 


95616 


95664 


95712 


95760 


95808 


95856 


91 


95904 


95951 


95999 


96047 


96094 


96142 


96189 


96231 i 


96284 


9(5331 


92 


96378 


96426 


96473 


96520 


96567 


96614 


96661 


9()708 


9(5754 


9(5801 


93 


96848 


96895 


9()941 


96988 


97034 


97081 


97127 


97174 


97220 


972(i6 


94 


97312 


97359 


97405 


97451 


97497 


97543 


97589 


97635 


97680 


97726 


95 


97772 


97818 


97863 


97909 


97954 


98000 


98045 


98091 


98136 


98181 


96 


98227 


98272 


98317 


98362 


98407 


98452 


98497 


98542 


98587 


98632 


97 


98677 


98721 


9876(5 


98811 


98855 


98900 


98945 


98989 


99033 


95)078 


98 


99122 


99166 


99211 


99255 


99299 


99343 


99387 


99431 


99475 


95)519 


99 


99563 


99607 


99651 


99(i94 


99738 


9i)782 


99825 


95)8(59 


99913 


95)5)5(5 



APPENDIX III 
TKIGONOMETRIC EUNCTIONS 



TRIGONOMETRIC FUNCTIONS 
1 



427 



NATURAL SINES 


», COSINES, TANGENTS. ETC. 


o 


/ 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 








.000000 


Infinite 


.000000 


Infinite 


1.00000 


1.000000 





90 




10 


.002909 


343.77516 


.002909 


343.77371 


1.00000 


.999996 


50 






20 


.005818 


171.88831 


.005818 


171.88540 


1.00002 


.999983 


40 






30 


.008727 


114.59301 


.008727 


114.58865 


1.00004 


.999962 


30 






40 


.011635 


85.945609 


.011636 


85.939791 


1.00007 


.999932 


20 






50 


.014544 


68.757360 


.014545 


68.750087 


1.00011 


.999894 


10 




1 





.017452 


57.298688 


.017455 


57.289962 


1.00015 


.999848 





89 




10 


.020361 


49.114062 


.020365 


49.103881 


1.00021 


.999793 


50 






20 


.023269 


42.975713 


.023275 


42.964077 


1.00027 


.999729 


40 






30 


.026177 


38.201550 


.026186 


38.188459 


1.00034 


.999657 


30 






40 


.029085 


34.382316 


.029097 


34.367771 


1.00042 


.999577 


20 






50 


.031992 


31.257577 


.032009 


31.241577 


1.00051 


.999488 


10 




2 





.034899 


28.653708 


.034921 


28.636253 


1.00061 


.999391 





88 




10 


.037806 


26.450510 


.037834 


26.431600 


1.00072 


.999285 


50 






20 


.040713 


24.562123 


.040747 


24.541758 


1.00083 


.999171 


40 






30 


.043619 


22.925586 


.043661 


22.903766 


1.00095 


.999048 


30 






40 


.046525 


21.493676 


.046576 


21.470401 


1.00108 


.998917 


20 






50 


.049431 


20.230284 


.049491 


20.205553 


1.00122 


.998778 


10 




3 





.052336 


19.107323 


.052408 


19.081137 


1.00137 


.998630 





87 




10 


.055241 


18.102619 


.055325 


18.074977 


1.00153 


.998473 


50 






20 


.058145 


17.1984:34 


.058243 


17.169337 


1.00169 


.998308 


40 






30 


.061049 


16.380408 


.061163 


16.349855 


1.00187 


.998135 


30 






40 


.063952 


15.636793 


.064083 


15.604784 


1.00205 


.997357 


20 






50 


.066854 


14.957882 


.067004 


14.924417 


1.00224 


.997763 


10 




4 





.069756 


14.335587 


.069927 


14.300666 


1.00244 


.997564 





86 




10 


.072658 


13.763115 


.072851 


13.726738 


1.00265 


.997357 


50 






20 


.075559 


13.234717 


.075776 


13.196888 


1.00287 


.997141 


40 






30 


.078459 


12.745495 


.078702 


12.706205 


1.00309 


.996917 


30 






40 


.081359 


12.291252 


.081629 


12.250505 


1.00333 


.996685 


20 






50 


.084258 


11.868370 


.084558 


11.826167 


1.00357 


.996444 


10 




5 





.087156 


11.473713 


.087489 


11430052 


1.00382 


.996195 





85 




10 


.090053 


11.104549 


.090421 


11.059431 


1.00408 


.995937 


50 






20 


.092950 


10.758488 


.093.354 


10.711913 


1.00435 


.995671 


40 






30 


.095846 


10.433431 


.096289 


10.38.5397 


1.00463 


.995396 


30 






40 


.098741 


10.127522 


.099226 


10.078031 


1.00491 


.995113 


20 






50 


.101635 


9.8391227 


.102164 


9.7881732 


1.00521 


.994822 


10 




6 





.104528 


9.5667722 


.105104 


9.5143645 


1.00551 


.994522 





84 




10 


.107421 


9.3091699 


.108046 


9.2553035 


1 .00582 


.994214 


50: 




20 


.110313 


9.0651512 


.1109CH) 


9.00982(51 


1.00614 


.993897 


40 83 


o 


/ 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine 


1 

/ 1 o 


For fiinc 


tions from ^?/ 


^ 40' to 90° 


read from bott 


oin of t:ib 


e upward. 



428 



TRIGONOMETRIC FUNCTIONS 



NATURAL SINES 


, COSINES, TANGENTS, ETC. 






(Continued) 






o 


/ 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


6 


30 


.113203 


8.8336715 


.113936 


8.7768874 


1.00647 


.993572 


30 






40 


.116093 


8.6137^)01 


.116883 


8.5555468 


1.00681 


.993238 


20 






50 


.118982 


8.4045586 


.119833 


8.3449558 


1.00715 


.992896 


10 




7 





.121869 


8.2055090 


.122785 


8.1443464 


1.00751 


.992546 





83 




10 


.124756 


8.0156450 


.125738 


7.9530224 


1.00787 


.992187 


50 






20 


.127642 


7.8:344335 


.128694 


7.7703506 


1.00825 


.991820 


40 






30 


.130526 


7.6612976 


.131653 


7.5957541 


1.008(53 


.991445 


30 






40 


.133410 


7.4957100 


.134613 


7.4287064 


1.00902 


.991061 


20 






50 


.136292 


7.3371909 


.137576 


7.2687255 


1.00942 


.990669 


10 




8 





.139173 


7.1852965 


.140541 


7.1153697 


1.00983 


.990268 





82 




10 


.142053 


7.0396220 


.143508 


6.9682335 


1.01024 


.989859 


50 






20 


.144932 


6.8997942 


.146478 


6.8269437 


1.01067 


.989442 


40 






30 


.147809 


6.7654691 


.149451 


6.6911562 


1.01111 


.989016 


30 






40 


.150686 


6.6363293 


.152426 


6.5605538 


1.01155 


.988582 


20 






50 


.153561 


6.5120812 


.155404 


6.4348428 


1.01200 


.988139 


10 




9 





.156434 


6.3924532 


.158384 


6.3137515 


1.01247 


.987688 





81 




10 


.159307 


6.2771933 


.161368 


6.1970279 


1.01294 


.987229 


50 






20 


.162178 


6.1660674 


.164354 


6.0844381 


1.01342 


.986762 


40 






30 


.165048 


6.0588980 


.167343 


5.9757644 


1.01391 


.986286 


30 






40 


.167916 


5.955.3625 


.170334 


5.8708042 


1.01440 


.985801 


20 






50 


.170783 


5.8553921 


.173329 


5.7693688 


1.01491 


.985309 


10 




10 





.173648 


5.7587705 


.176327 


5.6712818 


1.01543 


.984808 





80 




10 


.176512 


5.6653331 


.179328 


5.5763786 


1.01595 


.984298 


50 






20 


.179375 


5.5749258 


.182332 


5.4845052 


1.01649 


.983781 


40 






30 


.182236 


5.4874043 


.185,339 


5.3955172 


1.01703 


.983255 


30 






40 


.185095 


5.4026333 


.188359 


5.3092793 


1,01758 


.982721 


20 






50 


.187953 


5.3204860 


.191363 


5.2256647 


1.01815 


.982178 


10 




11 





.190809 


5.2408431 


.194380 


5.1445540 


1.01872 


.981627 





79 




10 


. 193664 


5.1635924 


.197401 


5.0658352 


1.01930 


.9810(58 


50 






20 


.196517 


5.0886284 


.200425 


4.9894027 


1.01989 


.980500 


40 






30 


.199368 


5.0158317 


.203452 


4.9151570 


1.02049 


.979925 


30 






40 


.202218 


4.9451687 


.206483 


4.8430045 


1.02110 


.979341 


20 






50 


.205065 


4.8764907 


.209518 


4.7728568 


1.02171 


.978748 


10 




12 





.207912 


4.8097343 


.212557 


4.7046.301 


1.022.34 


.978148 





78 




10 


.210756 


4.7448206 


.215.599 


4.(5382457 


1.02298 


.977539 


50 






20 


.213599 


4.6816748 


.218645 


4.. 573(5287 


1.02.3(52 


.976921 


40 






30 


.21()440 


4.(5202263 


.221695 


4.5107085 


1.02428 


.976296 


30 






40 


.219279 


4.5604080 


.224748 


4.4494181 


1.02494 


.975(562 


20 






50 


.222116 


4.5021565 


.227806 


4.3896940 


1.02562 


.975020 


10 


77 


o 


/ 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine ' 


o 


F 


or functions from 77° If 


' to 83° .30' read from b 


attorn of ts 


ible upward. 



TRIGONOMETBIC FUNCTIONS 



429 



NATURAL SINES 


, COSINES, TANGENTS, 


ETC. 






{Conti 


nued) 








o 


/ 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


13 





.224951 


4.4454115 


.230868 


4.3314759 


1.02630 


.974370 





77 




10 


.227784 


4.3901158 


.233934 


4.2747066 


1.02700 


.973712 


50 






20 


.230616 


4.3362150 


.237004 


4.2193318 


1.02770 


.973045 


40 






30 


.233445 


4.2836576 


.240079 


4.1652998 


1.02842 


.972370 


30 






40 


.236273 


4.2323943 


.243158 


4.1125614 


1.02914 


.971687 


20 






50 


.239098 


4.1823785 


.246241 


4.0610700 


1.02987 


.970995 


10 




14 





.241922 


4.1335655 


.249328 


4.0107809 


1.03061 


.970296 





76 




10 


.244743 


4.0859130 


.252420 


3.9616518 


1.03137 


.969588 


50 






20 


.247563 


4.0393804 


.255517 


3.9136420 


1.03213 


.968872 


40 






30 


.250380 


3.9939292 


.2.58618 


3.8667131 


1.032^)0 


.968148 


30 






40 


.253195 


3.9495224 


.261723 


3.8208281 


1.03363 


.967415 


20 






50 


.256008 


3.9061250 


.264834 


3.7759519 


1.03447 


.966675 


10 




15 





.258819 


3.8637033 


.267949 


3.7320508 


1.0.3528 


.965926 





75 




10 


.261628 


3.8222251 


.271069 


3.68tK)927 


1.03609 


.965169 


50 






20 


.264434 


3.7816596 


.274195 


3.6470467 


1.03691 


.964404 


40 






30 


.267238 


3.7419775 


.277325 


3.6058835 


1.03774 


.963630 


30 






40 


.270040 


3.7031506 


.280460 


3.5655749 


1.03858 


.962849 


20 






50 


.272840 


3.6651518 


.283600 


3.5260938 


1.03944 


.962059 


10 




16 





.275637 


3.6279553 


.286745 


3.4874144 


1.04030 


.961262 





74 




10 


.278432 


3.5915363 


.289896 


3.4495120 


1.04117 


.960456 


50 






20 


.281225 


3.5558710 


.293052 


3.4123626 


1.04206 


.959642 


40 






30 


.284015 


3.5209365 


.296214 


3.3759434 


1.04295 


.958820 


30 






40 


.286803 


3.4867110 


.299380 


3.3402326 


1.04385 


.957990 


20 






50 


.289589 


3.4531735 


.302553 


3.3052091 


1.04477 


.957151 


10 




17 





.292372 


3.4203036 


.305731 


3.2708526 


1.04569 


.956305 





73 




10 


.295152 


3.3880820 


.308914 


3.2371438 


1.04663 


.955450 


50 i 1 




20 


.297930 


3.3564^)00 


.312104 


3.2040638 


1.04757 


.954588 


40 






30 


.300706 


3.3255095 


.315299 


3.1715948 


1.04853 


.953717 


30 






40 


.303479 


3.29512.34 


.318500 


3.1397194 


1.04950 


.952838 


20 






50 


.306249 


3.2653149 


.321707 


3.1084210 


1.05047 


.951951 


10 




18 





.309017 


3.2.360680 


.324920 


3.0776835 


1.05146 


.951057 





72 




10 


.311782 


3.2073673 


..328139 


3.0474915 


1.05246 


.950154 


50 






20 


.314545 


3.1791978 


.331364 


3.0178301 


1.05347 


.949243 


40 






30 


.317305 


3.15154.53 


..334595 


2.9886850 


1.0.5449 


.948324 


30 






40 


.320062 


3.124.39.59 


.337833 


2.9600422 


1.05.5.52 


.947397 


20 






50 


.322816 


3.0977363 


.341077 


2.9318885 


1.05657 


.946462 


10 




19 





.325568 


3.07155.35 


.344.328 


2.9042109 


1.05762 


.945519 





71 




10 


.328317 


3.0458.352 


.347585 


2.8769970 


1.0.5869 


.944568 


50 






20 


.331063 


3.0205693 


.350848 


2.8502349 


1.05976 


.943609 


40 


70 





/ 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine 


/ 


o 


I 


""or functions from 70° 4 


0' to 77° 0' 


read from be 


)ttom of ta 


3le upward 





430 



TRIGONOMETRIC FUNCTIONS 



NATURAL 


SINES 


. COSINES, TANGENTS, 


ETC. 








(^Continued) 








o 


/ 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


19 


30 


.333807 


2.9957443 


.354119 


2.8239129 


1.06085 


.942641 


30 






40 


.336547 


2.9713490 


.357396 


2.7980198 


1.06195 


.941666 


20 






50 


.339285 


2.9473724 


.360680 


2.7725448 


1.06306 


.940684 


10 




20 





.342020 


2.9238044 


.363970 


2.7474774 


1.06418 


.939693 





70 




10 


.344752 


2.9006346 


.367268 


2.7228076 


1.06531 


.938694 


50 






20 


.347481 


2.8778532 


.370573 


2.6985254 


1.06645 


.9,37687 


40 






30 


.350207 


2.8554510 


.373885 


2.6746215 


1.06761 


.936672 


30 






40 


.352931 


2.8334185 


.377204 


2.6510867 


1.06878 


.935650 


20 






50 


.355651 


2.8117471 


.380530 


2.6279121 


1.06995 


.934619 


10 




21 





.358368 


2.7904281 


.383864 


2.6050891 


1.07115 


.933580 





69 




10 


.361082 


2.7694532 


.387205 


2.5826094 


1.07235 


.932534 


50 






20 


.363793 


2.7488144 


.390554 


2.5604649 


1.07356 


.931480 


40 






30 


.366501 


2.7285038 


.393911 


2.5386479 


1.07479 


.930418 


30 






40 


.369206 


2.7085139 


.397275 


2.5171507 


1.07602 


.929348 


20 






50 


.371908 


2.6888374 


.400647 


2.4959661 


1.07727 


.928270 


10 




22 





.374607 


2.6694672 


.404026 


2.4750869 


1.07853 


.927184 





68 




10 


.377302 


2.6503962 


.407414 


2.4545061 


1.07981 


.926090 


50 






20 


.379994 


2.6316180 


.410810 


2.4342172 


1.08109 


.924989 


40 






30 


.382683 


2.6131259 


.414214 


2.4142136 


1.08239 


.923880 


30 






40 


.385369 


2.5949137 


.417626 


2.3944889 


1.08370 


.922762 


20 






50 


.388052 


2.5769753 


.421046 


2.3750372 


1.08503 


.921638 


10 




23 





.390731 


2.5593047 


.424475 


2.3558,524 


1.086,36 


.920505 





67 




10 


.393407 


2.5418961 


.427912 


2.3369287 


1.08771 


.919364 


50 






20 


.396080 


2.5247440 


.431358 


2.3182606 


1.08907 


.918216 


40 






30 


.398749 


2.5078428 


.4134812 


2.2998425 


1.09044 


.917060 


30 






40 


.401415 


2.4911874 


.438276 


2.2816693 


1.09183 


.915896 


20 






50 


.404078 


2.4747726 


.441748 


2.2637357 


1.09323 


.914725 


10 




24 





.406737 


2.4585933 


.445229 


2.2460368 


1.09464 


.91,3545 





66 




10 


.409392 


2.442{)448 


.448719 


2.2285676 


1.09606 


.912358 


50 






20 


.412045 


2.4269222 


.452218 


2.2113234 


1.09750 


.911164 


40 






30 


.414693 


2.4114210 


.455726 


2.1942997 


1.09895 


.909961 


30 






40 


.417338 


2.3961367 


.459244 


2.1774920 


1.10041 


.908751 


20 






50 


.419980 


2.3810650 


.462771 


2.1608958 


1.10189 


.907533 


10 




25 





.422618 


2.3662016 


.466308 


2.1445069 


1.10,338 


.906,308 





65 




10 


.425253 


2.3515424 


.4()9854 


2.1283213 


1.10488 


.905075 


50 






20 


.427884 


2.3370833 


.473410 


2.112.3348 


1.10640 


.c)03834 


40 






30 


.430511 


2.3228205 


.47(5976 


2.0965436 


1.10793 


.902585 


30 






40 


.433135 


2.3087501 


.480551 


2.0809438 


1.10947 


.fK)1329 


20 






50 


.435755 


2.2948685 


.484137 


2.0655318 


1.11103 


.tX)0065 


10 


64 





1 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine 


t 


o 


I 


'or functioi 


IS from 64°-l 


J' to 70° -30' read from b 


ottom of tf 


ible upwarc 


1. 



TBIGONOMETRIC FUNCTIONS 



431 



NATURAL 


SINES 


, COSINES, TANGENTS, ETC. 








(^Continued) 




o 


/ 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


26 





.438371 


2.2811720 


.487733 


2.0.503038 


1.11260 


.898794 





64 




10 


.440984 


2.2676571 


.491339 


2.0352565 


1.11419 


.897515 


50 






20 


.443593 


2.2543204 


.494955 


2.0203862 


1.11579 


.896229 


40 






30 


.446198 


2.2411585 


.498582 


2.0056897 


1.11740 


.894934 


30 






40 


.448799 


2.2281681 


.502219 


1.9911637 


1.11903 


.89:3633 


20 






50 


.451397 


2.2153460 


.505867 


1.9768050 


1.12067 


.892323 


10 




27 





.453990 


2.2026893 


.509525 


1.9626105 


1.12233 


.891007 





63 




10 


.456580 


2.1901947 


.513195 


1.9485772 


1.12400 


.889682 


50 






20 


.459166 


2.1778595 


.516876 


1,9347020 


1.12568 


.888.350 


40 






;50 


.461749 


2.1656806 


.520567 


1.9209821 


1.12738 


.887011 


30 






40 


.464327 


2.153(i553 


.524270 


1.9074147 


1.12910 


.885664 


20 1 




50 


.466901 


2.1417808 


.527984 


1.8939971 


1.13083 


.884309 


10 




28 





.469472 


2.1300545 


.531709 


1.8807265 


1.13257 


.882948 





62 




10 


.472038 


2.1184737 


.535547 


1.8676003 


1.134:33 


.881578 


50 






20 


.474600 


2.1070359 


.539195 


1.8546159 


1.13610 


.880201 


40 






30 


.477159 


2.0957385 


.542956 


1.8417409 


1.1.3789 


.878817 


30 






40 


.479713 


2.0845792 


.546728 


1.8290628 


1.13970 


.877425 


20 






50 


.482263 


2.0735556 


.550515 


1.8164892 


1.14152 


.876026 


10 




29 





.484810 


2.0626653 


.554309 


1.8040478 


1.14335 


.874620 





61 




10 


.487352 


2.0519061 


.558118 


1.7917362 


1.14521 


.873206 


50 






20 


.489890 


2.0412757 


.561939 


1.7795524 


1.14707 


.871784 


40 






30 


.492424 


2.0307720 


.565773 


1.7674940 


1.14896 


.8703.56 


30 






40 


.494953 


2.0203929 


.569619 


1.7.555590 


1.15085 


.868920 


20 






50 


.497479 


2.0101362 


.573478 


1.7437453 


1.15277 


.867476 


10 




30 





.500000 


2.0000000 


.577350 


1.7320.508 


1.15470 


.866025 





60 


i 10 


.502517 


1.9899822 


.581235 


1.72047:36 


1.15665 


.864567 


50 






20 


.505030 


1.9800810 


.5851:34 


1.7090116 


1.15861 


.863102 


40 






30 


.507538 


1.9702944 


.589045 


1.6976631 


1.16059 


.861629 


30 






40 


.510043 


1.9606206 


.592970 


1.6864261 


1.16259 


.860149 


20 






50 


.512543 


1.9510577 


.596908 


1.6752988 


1.16460 


.858662 


10 




31 





.515038 


1.9416040 


.600861 


1.6642795 


1.16663 


.857167 





59 




10 


.517529 


1.9322.-)78 


.604827 


1.653:3663 


1.16868 


.855665 


50 






20 


.520016 


1.9230173 


.608807 


1.6425576 


1.17075 


.854156 


40 






30 


.522499 


1.9138809 


.612801 


1.6:318517 


1.17283 


.852640 


30 






40 


.524977 


1.9048469 


.616809 


1.6212469 i 1.17493 


.8.51117 


20 




50 


.527450 


1.8959138 


.620832 


1.6107417 


1.17704 


.849586 


10 




32 


.529919 


1.8870799 


.624869 


1.600.3.345 


1.17918 


.848048 





58 


10 


.532384 


1.878:3438 


.628921 


1.5900238 


1.18i:33 


.846503 


50 ' 


20 


.534844 


1.8697040 


.632988 


1.5798079 


1.18350 


.844951 


40 57 


o 


/ 


Cosine 


Sei-ant 


Cotangent 


Tangent 


Cosecant 


Sine 


/ 


o 




^'or functii 


ns from 57°- 


to' to 04°-0' re;i(l from bottom of ta 


ble upward. 



2 A 



432 



TBIGONOMETRIC FUNCTIONS 



6 



NATURAL SINES 


, COSINES, TANGENTS, 


ETC. 






(Cont 


inued) 








o 


/ 


Sine 


Cosecant 


Tano:ent 


Cotangent 


Secant 


Cosine 


/ 


o 


32 


30 


.537300 


1.8611590 


.637079 


1.5696856 


1.18569 


.843391 


30 






40 


.539751 


1.8527073 


.641167 


1.5596552 


1.18790 


.841825 


20 






50 


.542197 


1.8443476 


.645280 


1.5497155 


1.19012 


.840251 


10 




33 





.544639 


1.8360785 


.649408 


1.5398650 


1.19236 


.838671 





57 




10 


.547076 


1.8278985 


.653531 


1.5301025 


1.19463 


.837083 


50 






20 


.549509 


1.8198065 


.657710 


1.5204261 


1.19691 


.835488 


40 






30 


.551937 


1.8118010 


.661886 


1.5108352 


1.19920 


.833886 


30 






40 


.554360 


1.8038809 


.666077 


1.5013282 


1.20152 


.832277 


20 






50 


.556779 


1.7960449 


.670285 


1.4919039 


1.20386 


830661 


10 




34 





.559193 


1.7882916 


.674509 


1.4825610 


1.20622 


.829038 





56 




10 


.561602 


1.7806201 


.678749 


1.4732983 


1.20859 


.827407 


50 






20 


.564007 


1.7730290 


.683007 


1.4641147 


1.21099 


.825770 


40 






30 


.566406 


1.7655173 


.687281 


1.4550090 


1.21341 


.824126 


30 






40 


.568801 


1.7580837 


.691573 


1.4459801 


1.21584 


.822475 


20 






50 


.571191 


1.7507273 


.695881 


1.4370268 


1.21830 


.820817 


10 




35 





.573576 


1.7434468 


.700208 


1.4281480 


1.22077- 


.819152 





55 




10 


.575957 


1.7362413 


.704552 


1.4193427 


1.22327 


.817480 


50 






20 


.578332 


1.7291096 


.708913 


1.4106098 


1.22579 


.815801 


40 






30 


.580703 


1.7220508 


.713293 


1.4019483 


1.22833 


.814116 


30 






40 


.583069 


1.7150639 


.717691 


1,3933571 


1.23089 


.812423 


20 






50 


.585429 


1.7081478 


.722108 


1.3848355 


1.23347 


.810723 


10 




36 





.587785 


1.7013016 


.726543 


1.3763810 


1.23607 


.809017 





54 




10 


.590136 


1.6945244 


.730996 


1.3679959 


1.23869 


.807304 


50 






20 


.592482 


1.6878151 


.735469 


1.359(5764 


1.24134 


.805584 


40 






30 


.594823 


1.6811730 


.739961 


1.3514224 


1.24400 


.803857 


30 






40 


.597159 


1.6745970 


.744472 


1.3432331 


1.24669 


.802123 


20 






50 


.699489 


1.6680864 


.749003 


1.3351075 


1.24940 


.800383 


10 




37 





.601815 


1.6616401 


.753554 


1.. 3270448 


1.25214 


.798636 





53 




10 


.604136 


1.6552575 


.758125 


1.3190441 


1.25489 


.796882 


50 






20 


.606451 


1.6489376 


.762716 


1.3111046 


1.25767 


.795121 


40 






30 


.608761 


1.6426796 


.767627 


1.3032254 


1.26047 


.793353 


30 






40 


.611067 


1.6364828 


.771959 


1.2954057 


1.26330 


.791579 


20 






50 


.613367 


1.6303462 


.776612 


1.2876447 


1.26615 


.789798 


10 




38 





.615661 


1.6242692 


.781286 


1.2799416 


1.26902 


.788011 





52 




10 


.617951 1 1.6182510 


.785981 


1.27229.57 


1.27191 


.786217 


50 






20 


.620235 


1.6122908 


,790698 


1.2647062 


1.27483 


.784416 


40 






30 


.622515 


1.6063879 


.795436 


1.2571723 


1.27778 


.782608 


30 






40 


.624789 


1.6005416 


.800196 


1.2496933 


1.28075 


.780794 


20 






50 


.627057 


1.5947511 


.804080 


1.2422685 


1.28374 


.778973 


10 


51 


o 


/ 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine ' 


o 


I 


"■or functions from 51°-1 


0' to 5T°-3C 


' read from b 


ottoin of ti 


ible upwar 


1. 



TRIGONOMETRIC FUNCTIONS 



433 



NATURAL 


SINES 


, COSINES, TANGENTS, ETC. 








{Continued) 







t 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


39 





.629320 


1.5890157 


?809784 


1.2348972 1.28676 


.777146 





51 




10 


.631578 


1.5833318 


.814612 


1.2275786 


1.28980 


.775312 


50 






20 


.633831 


1.5777077 


.819463 


1.2203121 


1.29287 


.773472 


40 






30 


.636078 


1.5721337 


.824336 


1.2130970 


1.29597 


.771625 


30 






40 


.638320 


1.5666121 


.829234 


1.2059327 


1.29909 


.769771 


20 






50 


.640557 


1.5611424 


.834155 


1.1988184 


1.30223 


.767911 


10 




40 





.642788 


1.5557238 


.839100 


1.1917536 


1.30541 


.766044 





50 




10 


.645013 


1.5503558 


.844069 


1.1847376 


1.30861 


.764171 


50 






20 


.647233 


1.5450378 


.849062 


1.1777698 


1.31183 


.762292 


40 




i 30 


.649448 


1.5397690 


.854081 


1.1708496 


1.31509 


.760406 


30 






40 


.651657 


1.5345491 


.859124 


1.1639763 


1.31837 


.758514 


20 






50 


.653861 


1.5293773 


.864193 


1.1571495 


1.32168 


.756615 


10 




41 





.656059 


1.5242531 


.869287 


1.1503684 


1.32501 


.754710 





49 




10 


.658252 


1.5191759 


.874407 


1.1436326 


1.32838 


.752798 


50 






20 


.660439 


1.5141452 


.879553 


1.1369414 


1.33177 


.750880 


40 






30 


.662620 


1.5091605 


.884725 


1.1302944 


1.33519 


.748956 


30 






40 


.664796 


1.5042211 


.889924 


1.1236909 


1.33864 


.747025 


20 






50 


.666966 


1.4993267 


.895151 


1.1171305 


1.34212 


.745088 


10 




42 





.669131 


1.4944765 


.900404 


1.1106125 


1.34563 


.743145 





48 




10 


.671289 


1.4896703 


.905685 


1.1041365 


1.34917 


.741195 


50 






20 


.673443 


1.4849073 


.910994 


1.0977020 


1.35274 


.739239 


40 






30 


.675590 


1.4801872 


.916331 


1.0913085 


1.35634 


.737277 


30 






40 


.677732 


1.4755095 


.921697 


1.0849554 


1.35997 


.735309 


20 






50 


.679868 


1.4708736 


.927091 


1.0786423 


1.36363 


.733335 


10 




43 





.681998 


1.4662792 


.932515 


1.0723687 


1.36733 


.731354 





47 




10 


.684123 


1.4617257 


.937968 


1.0661341 


1.37105 


.729367 


50 






20 


.686242 


1.4572127 


.943451 


1.0599381 


1.37481 


.727374 


40 






30 


.688355 


1.4527397 


.948965 


1.0537801 


1.37860 


.725374 


30 






40 


.690462 


1.4483063 


.954508 


1.0476598 


1.38242 


.723369 


20 






50 


.692563 


1.4439120 


.960083 


1.0415767 


1.38628 


.721357 


10 




44 





.694658 


1.4395565 


.965689 


1.0355303 


1.39016 


.719340 





46 




10 


.696748 


1.4352393 


.971326 


1.0295203 


1 .39409 


.717316 


50 






20 


.698832 


1.4309602 


.976996 


1.0235461 


1.39804 


.715286 


40 




30 


.700f)09 


1.4267182 


.982697 


1.017()074 


1.40203 


.713251 


30 




' 40 


.702981 


1.4225i;34 


.988432 


1.0117088 


1.40606 


.711209 


20 




50 


.705047 


1.4183454 


.994199 


1.0058348 


1.41012 


.709161 


10 




45 


.707107 


1.4142136 


1.000000 


1.0000000 


1.41421 


.707107 





45 


o 1 1 

1 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine 


1 


o 




For functic 


)ns from 45°- 


3' to 51°-0' read from bottom of tab 


le upward. 



APPENDIX IV 

CONVERSION TABLES 



CONVERSION TABLES 



437 



TABLES 


FOR CONVERTING UNITED STATES 






WEIGHTS ANE 


> MEASURES 








METRIC TO CUSTOMARY 








WEIGHTS 






Milligrams 


firams 


Grams 


Kilograms 


Tonnes to 


Tonnes to 


No. 


to 


to 


to Avoirdupois 


to Avoirdupois 


Net Tons of 


Gross Tons of 




tiraiiis 


Troy Ounces 


Ounces 


Pounds 


2000 Pounds 


2240 Pounds 


1 


.01543 


.03215 


.03527 


2.20462 


1.10231 


.98421 


2 


.03086 


•06430 


.07055 


4.40924 


2.20462 


1.96841 


3 


.04630 


.09645 


.10582 


6.61387 


3.30693 


2.95262 


4 


.06173 


.12860 


.14110 


8.81849 


4.40924 


3.93682 


5 


.07716 


.16075 


.17637 


11.02311 


5.51156 


4.92103 


6 


.09259 


.19290 


.21164 


13.22773 


6.61387 


5.90524 


7 


.10803 


.22506 


.24692 


15.4.3236 


7.71618 


6.88944 


8 


.12346 


.25721 


.28219 


17.63698 


8.81849 


7.87365 


9 


.13889 


.28936 


.31747 


19.84160 


9.92080 


8.85785 






1 Kilogram = 1543 


2.35639 Grains 








LINEAR Ml 


EASURE 






Millimeters 


Centimeters 


Meters 


Meters 


Kilomeiters 


Kilometers 


No. 


to 64tlis of an 


to 


to 


to 


to 


to 




Inch 


Iiulies 


Feet 


Yards 


Statute Miles 


Nautical Miles 


1 


2.51968 


.39370 


3.280833 


1.093611 


.62137 


.53959 


2 


5.03936 


.78740 


6.561667 


2.187222 


1.24274 


1.07919 


3 


7.55904 


1.18110 


9.842500 


3.280833 


1.86411 


1.61878 


4 


10.07872 


1,57480 


13.12.3333 


4.374444 


2.48548 


2.15837 


5 


12.59840 


1.968.50 


16.404167 


5.468056 


3.10685 


2.69796 


6 


15.11808 


2.36220 


19.685000 


6.561667 


3.72822 


3.23756 


7 


17.63776 


2.75590 


22.965833 


7.655278 


4.34959 


3.77715 


8 


20.15744 


3.14960 


26.246667 


8.748889 


4.97096 


4.31674 


9 


22.67712 


3.54330 


29.527.500 


9.842.500 


5.59233 


4.85633 





438 



CONVERSION TABLES 



TABLES 


FOR CONVERTING UNITED STATES 






^WEIGHTS AND 


MEASURES 








CUSTOMARY TO METRIC 










WEIGHTS 






Giauis 


. Troy Ounces 


Avoirdupois 


Avoirdupois 


Net Tons of 


Gross Tons of 


l\0. 


to 


to 


Ounces 


Pounds to 


2000 Pounds 


2240 Pounds 




Milligrams 


Grams 


to Grams 


Kiloi^rams 


to Tonnes 


to Ton:: OS 


1 


64.79892 


31.10348 


28.34953 


.45359 


.90718 


1 .01605 


2 


129.59784 


62.20696 


56.69905 


mi\% 


1.81437 


2.03209 


3 


194.39675 


93.31044 


85.04858 


1.36078 


2.72155 


3.04814 


4 


259.19567 


124.41392 


113.39811 


1.81437 


3.(i2874 


4.06419 


5 


323.99459 


155.51740 


141.74763 


2.26796 


4.5.3592 


5.08024 


() 


, 388.79351 


186.62088 


170.09716 


2.72155 


5.44311 


6.09628 


7 


453.59243 


217.72437 


198.44669 


3.17515 


6.35029 


7.11233 


8 


518.39135 


248.82785 


226.79621 


3.62874 


7.2.5748 


8.12838 


9 


583.19026 


279.93133 


255.14574 


4.08233 


8.16466 


9.14442 






1 Avoirdup 
L 


ois Pound = 
.INEAR Ml 


453.5924277 Grams 
EASURE 






64tlis of an 


Inches 


Feet 


Yards 


Statute Miles 


Nautical Miles 


No. 


Inch to 


to 


to 


to 


to 


to 




Millimeters 


Centimeters 


Meters 


Meters 


Kilometers 


Kilometers 


1 


.39688 


2.54001 


.304801 


.914402 


1.60935 


1.85.325 


2 


.79375 


5.08001 


.609601 


1.828804 


3.21869 


3.70650 


3 


1.19063 


7.62002 


.914402 


2.743205 


4.82804 


5.55975 


4 


1.58750 


10.16002 


1.219202 


3.657607 


6.43739 


7.41300 


5 


1.98438 


12.70003 


1.524003 


4.572009 


8.04674 


9.26625 


6 


2.38125 


15.24003 


1.828804 


5.486411 


9.65608 


11.11950 


7 


2.77813 


17.78004 


2.133604 


6.400813 


11.26543 


12.97275 


8 


3.17501 


20.32004 


2.4.38405 


7.315215 


12.87478 


14.82600 


9 


3.57188 


22.86005 


2.743205 


8.229616 


14.48412 


16.67925 






1 Nautit 


;al Mile = 


1853.25 Meters 








1 Gunte 


r's Chaiu = 


20.1168 Meters 








1 Fathoi 


11 = 


1 .829 Meters 





INDEX 



The numbers refer to the pages. 



Absorption dynamometer, 323. 
Acceleration, 184. 

angular, 232. 

components of, 208. 

constant, 185. 

in a curved path, 205. 

of points of a body having plane 
motion, 237. 

variable, 191. 
Angular momentum, 369. 

vector representation of, 373. 
Angular velocity, 232, 235. 

vector representation of, 241. 
Antifriction wheels, 298. 
Arch, linked, 158. 

masonry, 159. 

Balance, standing and numing of a 

shaft, 359. 
Ball bearings, 302. 
Bending moments, 161. 

diagram of, 162. 

graphical construction of, 162. 
Bow's notation, 99. 
Brake shoe testing machine, 272. 

Car on single rail, 382. 

Catenary, 177. 

Center of gravity, 38, 39, 43. 

by graphical methods, 57. 

by integration, 44. 

by Simpson's rule, 65. 

motion of, 377. 

of counterbalance, 56. 
Center of instantaneous rotation, 405. 
Center of oscillation, 345. 
Center of percussion, 405. 
Center of suspension, 345. 
Coefficient of friction, 247, 283. 

of belting, 316. 
Coefficient of restitution, 393. 
CoejSicient of rolling friction, 296. 



Compound pendulum, 344. 
Compression, 14. 
Concurrent forces, 10, 19. 
Connecting rod, 366. 
Conservation of energy, 254. 
Cords and pulleys, 170. 
Couples, 27, 71. 

combination of, 72, 76. 

components of, 78. 

equivalent, 73. 

gyroscopic, 380. 

moment of, 71, 75. 

substitution of force and couple 
for a force, 80. 

vector representation of, 77. 
Creeping of belts, 315. 

D'Alembert's principle, 334. 
Determination of g, 347. 
Displacement, 3. 
Durand's rule, 68. 
Dynamometer, absorption, 323. 
transmission, 314. 

Effective forces, 267, 334. 
Efficiency of wedge, 288. 
Elasticity of material, 397. 

modulus of, 398. 
Ellipse of inertia, 130. 
Ellipsoid of inertia, 140. 
Energy, 254. 
Equilibrant, 94. 

graphical method of finding, 94. 
Equilibrium, 1. 

conditions of, 12, 20, 29, 83, 92. 

of three forces, 16. 

of two forces, 14. 

Falling bodies, 186. 
Flexible cords, 148. 

as a catenary, 177. 

as a parabola, 171. 



439 



440 



INDEX 



Force, 1. 

transmissibility of, 6. 

unit of, 2. 

vector representation of, 5. 
Forces, concurrent, 6, 10, 19. 

in space, 90. 

non-concurrent, 82. 

polygon of, 8. 

resolution of, 7. 

triangle of, 7. 
Foucault's pendulum, 243. 
Friction brake, 324. 

Prony, 326. 
Friction circle, 317. 
Friction, coefficient of, 247, 283. 

laws of, for dry surfaces, 285. 

of belts, 310. 

of brake shoe, 328. 

of lubricated surfaces, 288. 

of pivots, 318. 

of worn bearing, 316. 

rolling, 296. 
Friction gears, 307. 

Guldinus, theorem of, 68. 
Gyroscope, 379. 

inclined axis, 383. 
Gyroscopic couple, 380. 

Harmonic motion, 192. 
Helical chute, 229. 
Horse power, 253. 
Hyperbolic functions, 180. 
tables of, appendix. 

Impact, direct, oblique, central, 390. 

direct central, 391. 

direct eccentric, 401. 

oblique against smooth plane, 408. 

of rotating bodies, 411. 

on a body with fixed axis, 403. 
Impact tension and compression, 398. 
Impressed forces, 266, 334. 
Impulse of a force, 394. 
Inertia, 1. 

ellipse of, 130. 

ellipsoid of, 140. 

moment of, 106. 

product of, 112. 
Instantaneous axis of rotation, 238,407. 



center of rotation, 239, 407. 
Internal stress couple, 166. 

Kinetic energy, 254. 
lost in impact, 395. 
of a body having plane motion, 274. 
of a body with one fixed point, 373. 
of rolling bodies, 279. 

Line of resistance of arch, 159. 
Lubricants, testing of, 293. 

Mass, 1. 

unit of, 2, 188. 

engineer's unit of, 188. 
Method of substitution, 103. 
Modulus of elasticity, 167, 398. 
Moment of a force, 22, 24. 

of the resultant, 23, 25, 28. 
Moment of inertia, 106, 268. 

axes of greatest and least, 112, 128. 

by direct addition, 126. 

by experiment, 346, 351. 

by graphical method, 124. 

by parallel sections, 135. 

by Simpson's rule, 126. 

polar, 115. 

principal moments of inertia, 139. 

theorem of inclined axes for. 111. 

theorem of parallel axes for, 109. 

units of, 108. 
Moment of momentum, 369. 

of a body with one fixed point, 371. 

rate of change of, due to rotating 
axes, 375. 
Momentum of a body, 394. 
Motion, along a curve in a vertical 
plane, 211. 

curvilinear, 204. 

in a circle, 233. 

in a twisted curve, 227. 

of center of gravity of body, 377. 

on inclined plane, 189. 

rotary, 232. 

uniform, in a circle, 209. 

where attraction varies directly as 

distance squared, 197. 
where resistance varies as distance, 

194. 
with repulsive force, 194. 



INDEX 



441 



Neutral surface, 168. 
Newton's laws of motion, 187. 

Pappus, theorem of, 68. 

Parabola, 171. 

Parallel forces, center of, 32, 35. 

resultant of, 27, 29. 
Pendulum, constant of, 398. 

compound, 344. 

conical, 209. 

cycloidal, 219. 

simple, 214. 
Period, of compound pendulum, 345. 

of deformation, 391. 

of restitution, 391. 

of simple pendulum, 216. 
Piledriver, 259. 
Pivots, flat end, 318. 

collar bearing, 318. 

conical, 319. 

Schiele's, 322. 

spherical, 321. 
Polar moment of inertia, 115. 
Pole of string polygon, 154. 
Power, transmitted by a belt, 313. 

transmitted by friction wheel, 309. 
Precession, 380. 

steady, 384. 

unsteady, 386. 
Principal axes, 113, 139. 
Principal moments of inertia, 139. 
Product of inertia, 112. 
Projectile in vacuo, 222. 

in resisting medium, 226. 

Radius of gyration, 107. 

Reaction of supports of rotating 

body, 340. 
Relative velocity, 200. 
Resistance of roads, 299. 
Rigid body, 4. 
Roller bearings, 301. 
Rolling friction, 296. 
Rotation about a fixed axis, 266, 336. 

with constant angular velocity, 353. 
Rotation and translation, 365. 
Rotation, of flywheel of steam en- 
gine, 362. 

of locomotive drive wheel, 358. 



of the earth, 243. 
under no forces, 352. 

Shear, 166. 
Simpson's rule, 62. 

applications of, 64. 
Simultaneous rotation about inter- 
secting axes, 240. 
Specific gravity, table of, 3. 
Speed, 183. 
Steam hammer, 262. 
Stresses in beams, 166. 
Stresses in frames, 99. 
String polygon, 153. 

depth of, 156. 

locus of pole of, 154. 
Substitution, method of, 103. 

Tension, 14. 

Torque and angular momentum, 370. 

Torsion balance, 348. 

Train resistance, 330. 

Translation of rigid body, 335. 

Transmission dynamometer, 314. 

Unit strain, 167. 
Unit stress, 167. 
Unit weights, table of, 3. 

Varignon's theorem, 23. 
Vector rate of change due to rota- 
tion, 374. 
Vector representation of angular 

momentum, 373. 
Vectors, 3. 
Velocity, 183. 
of spin, 383. 
relative, 200. 

Weight, 3. 

unit of, 3. 
Weights suspended by cords, 150. 

graphical solution of, 151. 
Work and energy, 245, 255, 276, 279. 

units of, 248. 
Work of a variable force, 250. 

of components of a force, 249. 

graphical representation of, 251. 

in uniform motion, 280. 



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A Textbook of 

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Chapter 


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rV. Molding Exercises without 


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The Elements of Electrical Transmission 

A Text-book for Colleges and Technical Schools 
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miiiiiiiH 



LIBRARY OF CONGRESS 



028 



20 1 



53 



